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Chapter 4
Random Variables
and Probability Distributions
4.1
4.2
a.
The number of newspapers sold by New York Times each month can take on a countable number of
values. Thus, this is a discrete random variable.
b.
The amount of ink used in printing the Sunday edition of the New York Times can take on an infinite
number of different values. Thus, this is a continuous random variable.
c.
The actual number of ounces in a one gallon bottle of laundry detergent can take on an infinite
number of different values. Thus, this is a continuous random variable.
d.
The number of defective parts in a shipment of nuts and bolts can take on a countable number of
values. Thus, this is a discrete random variable.
e.
The number of people collecting unemployment insurance each month can take on a countable
number of values. Thus, this is a discrete random variable.
a.
The closing price of a particular stock on the New York Stock Exchange is discrete. It can take on
only a countable number of values.
b.
The number of shares of a particular stock that are traded on a particular day is discrete. It can take
on only a countable number of values.
c.
The quarterly earnings of a particular firm is discrete. It can take on only a countable number of
values.
d.
The percentage change in yearly earnings between 2008 and 2009 for a particular firm is continuous.
It can take on any value in an interval.
e.
The number of new products introduced per year by a firm is discrete. It can take on only a
countable number of values.
f.
The time until a pharmaceutical company gains approval from the U.S. Food and Drug
Administration to market a new drug is continuous. It can take on any value in an interval of time.
4.3
Since there are only a fixed number of outcomes to the experiment, the random variable, x, the number of
stars in the rating, is discrete.
4.4
The number of customers, x, waiting in line can take on values 0, 1, 2, 3, … . Even though the list is never
ending, we call this list countable. Thus, the random variable is discrete.
4.5
The variable x, total compensation in 2008 (in $ millions), is reported in whole number dollars. Since there
are a countable number of possible outcomes, this variable is discrete.
4.6
A banker might be interested in the number of new accounts opened in a month, or the number of
mortgages it currently has, both of which are discrete random variables.
152
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
153
4.7
An economist might be interested in the percentage of the work force that is unemployed, or the current
inflation rate, both of which are continuous random variables.
4.8
The manager of a hotel might be concerned with the number of employees on duty at a specific time, or the
number of vacancies there are on a certain night.
4.9
The manager of a clothing store might be concerned with the number of employees on duty at a specific
time of day, or the number of articles of a particular type of clothing that are on hand.
4.10
A stockbroker might be interested in the length of time until the stockmarket is closed for the day.
4.11
a.
When a die is tossed, the number of spots observed on the upturned face can be 1, 2, 3, 4, 5, or 6.
Since the six sample points are equally likely, each one has a probability of 1/6.
The probability distribution of x may be summarized in tabular form:
4.12
x
1
2
3
4
5
6
p(x)
1
6
1
6
1
6
1
6
1
6
1
6
b.
The probability distribution of x may
also be presented in graphical form:
a.
The variable x can take on values 1, 3, 5, 7, and 9.
b.
The value of x that has the highest probability associated with it is 5. It has a probability of .4.
c.
Using MINITAB, the probability
distribution of x as a graph is:
d.
P(x = 7) = .2
e.
P(x ≥ 5) = p(5) + p(7) + p(9) = .4 + .2 + .1 = .7
f.
P(x > 2) = p(3) + p(5) + p(7) + p(9) = .2 + .4 + .2 + .1 = .9
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
154
4.13
Chapter 4
a.
We know
p( x) = 1. Thus, p(2) + p(3) + p(5) + p(8) + p(10) = 1
p(5) = 1 p(2) p(3) p(8) p(10) = 1 .15 .10 .25 .25 = .25
4.14
b.
P(x = 2 or x = 10) = P(x = 2) + P(x = 10) = .15 + .25 = .40
c.
P(x ≤ 8) = P(x = 2) + P(x = 3) + P(x = 5) + P(x = 8) = .15 + .10 + .25 + .25 = .75
a.
This is not a valid distribution because
b.
This is a valid distribution because 0 ≤ p(x) ≤ 1 for all values of x and
c.
This is not a valid distribution because p(4) = .3 < 0.
d.
The sum of the probabilities over all possible values of the random variable is
p( x) = .9 ≠ 1.
p( x) = 1.
p( x) = 1.1 > 1, so
this is not a valid probability distribution.
4.15
a.
The sample points are (where H = head, T = tail):
HHH HHT HTH THH HTT THT TTH TTT
3
2
2
2
1
1
1
0
1 1
If each event is equally likely, then P(sample point) =
n 8
x = # heads
b.
p(3) =
4.16
1
1 1 1 3
1 1 1 3
1
, p(2) = , p(1) = , and p(0) =
8
8 8 8 8
8 8 8 8
8
c.
Using Minitab, the graph of p(x) is:
d.
P(x = 2 or x = 3) = p(2) + p(3) =
a.
μ = E ( x) =
3 1 4 1
8 8 8 2
xp( x)
= 10(.05) + 20(.20) + 30(.30) + 40(.25) + 50(.10) + 60(.10)
= .5 + 4 + 9 + 10 + 5 + 6 = 34.5
σ2 = E(x − μ)2 =
σ =
(x μ )
2
p ( x)
= (10 34.5)2(.05) + (20 34.5)2(.20) + (30 34.5)2(.30)
+ (40 34.5)2(.25) + (50 34.5)2(.10) + (60 34.5)2(.10)
= 30.0125 + 42.05 + 6.075 + 7.5625 + 24.025 + 65.025 = 174.75
174.75 = 13.219
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
b.
c.
μ ± 2σ 34.5 ± 2(13.219) 34.5 ± 26.438 (8.062, 60.938)
P(8.062 < x < 60.938) = p(10) + p(20) + p(30) + p(40) + p(50) + p(60)
= .05 + .20 + .30 + .25 + .10 + .10 = 1.00
4.17
a.
μ = E(x) =
xp( x) = 4(.02) + (3)(.07) + (−2)(.10) + (−1)(.15) + 0(.3)
+ 1(.18) + 2(.10) + 3(.06) + 4(.02)
= −.08 − .21 − .2 − .15 + 0 + .18 + .2 + .18 + .08 = 0
σ2 = E[(x − μ)2] =
σ=
( x μ ) p( x)
2
= (−4 − 0)2(.02) + (−3 − 0)2(.07) + (−2 − 0)2(.10)
+ (−1 − 0)2(.15) + (0 − 0)2(.30) + (1 − 0)2(.18)
+ (2 − 0)2(.10) + (3 − 0)2(.06) + (4 − 0)2(.02)
= .32 + .63 + .4 + .15 + 0 + .18 + .4 + .54 + .32 = 2.94
2.94 = 1.715
b.
μ ± 2σ 0 ± 2(1.715) 0 ± 3.430 (−3.430, 3.430)
c.
P(−3.430 < x < 3.430) = p(−3) + p(−2) + p(−1) + p(0) + p(1) + p(2) + p(3)
= .07 + .10 + .15 + .30 + .18 + .10 + .06 = .96
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
155
156
4.18
Chapter 4
a.
It would seem that the mean of both would be 1 since they both are symmetric distributions centered
at 1.
b.
P(x) seems more variable since there appears to be greater probability for the two extreme values of 0
and 2 than there is in the distribution of y.
c.
For x: μ = E(x) =
σ2
xp( x) = 0(.3) + 1(.4) + 2(.3) = 0 + .4 + .6 = 1
= E[(x − μ) ] = ( x μ ) p ( x)
2
2
= (0 − 1)2(.3) + (1 − 1)2(.4) + (2 − 1)2(.3) = .3 + 0 + .3 = .6
yp( y) = 0(.1) + 1(.8) + 2(.1) = 0 + .8 + .2 = 1
= E[(y − μ) ] = ( y μ ) p( y )
For y: μ = E(y) =
σ2
2
2
= (0 − 1)2(.1) + (1 − 1)2(.8) + (2 − 1)2(.1) = .1 + 0 + .1 = .2
The variance for x is larger than that for y.
4.19
a.
The probability distribution for x is found by converting the Percent column to a probability column
by dividing the percents by 100. The probability distribution of x is:
x
2
3
4
5
b.
P(x = 5) = p(5) = .1837.
c.
P(x ≤ 2) = p(2) = .0408.
d.
μ E ( x)
p(x)
.0408
.1735
.6020
.1837
4
x p( x ) 2(.0408) 3(.1735) 4(.6020) 5(.1837)
i
i
i 1
.0816 .5205 2.4080 .9185 3.9286 3.93
The average star rating for a car’s drivers-side star rating is 3.9286.
4.20
a.
Yes. Relative frequencies are observed values from a sample. Relative frequencies are commonly
used to estimate unknown probabilities. In addition, relative frequencies have the same properties as
the probabilities in a probability distribution, namely
1. all relative frequencies are greater than or equal to zero
2. the sum of all the relative frequencies is 1
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
b.
157
Using MINITAB, the graph of the probability distribution is:
0.15
p(age)
0.10
0.05
0.00
20
25
30
age
c.
Let x = age of employee. Then P(x > 30) = .13 + .15 + .12 = .40.
P(x > 40) = 0
P(x < 30) = .02 + .04 + .05 + .07 + .04 + .02 + .07 + .02 + .11 + .07 = .51
4.21
d.
P(x = 25 or x = 26) = .02 + .07 = .09
a.
Yes. For all values of x, 0 ≤ p(x) ≤ 1 and
p( x) = .01 + .02 + .03 + .05 + .08 + .09 + .11 + .13 +
.12 + .10 + .08 + .06 + .05 + .03 + .02 + .01 + .01 = 1.00.
4.22
4.23
b.
P(x = 16) = .06
c.
P(x ≤ 10) = p(5) + p(6) + p(7) + p(8) + p(9) + p(10)
= .01 + .02 + .03 + .05 + .08 + .09 = .28
d.
P(5 ≤ x ≤ 15) = p(5) + p(6) + p(7) + p(8) + p(9) + p(10) + p(11) + p(12) + p(13) + p(14) + p(15)
= .01 + .02 + .03 + .05 + .08 + .09 + .11 + .13 + .12 + .10 + .08
= .82
a.
The probability distribution for x is:
Grill Display
Combination
1-2-3
x
6
p(x)
35 / 124 = .282
1-2-4
7
8 / 124 = .065
1-2-5
8
42 / 124 = .339
2-3-4
9
4 / 124 = .032
2-3-5
10
1 / 124 = .008
2-4-5
11
34 / 124 = .274
b.
P(x > 10) = p(10) + p(11) = .008 + .274 = .282
a.
X is a discrete random variable because it can take on only values 0, 1, 2, 3, 4, or 5 in this
example.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
158
Chapter 4
b.
c.
p(0)
5!(.35)0 (.65)5 0 5 4 3 2 1(1)(.65)5
.655 .1160
0!(5 0)!
1 5 4 3 2 1
p(1)
5!(.35)1 (.65)51 5 4 3 2 1(.35)1 (.65) 4
5(.35)(.65) 4 .3124
1!(5 1)!
1 4 3 2 1
p(2)
5!(.35)2 (.65)5 2 5 4 3 2 1(.35)2 (.65)3
10(.35)2 (.65)3 .3364
2!(5 2)!
2 1 3 2 1
p(3)
5!(.35)3 (.65)53 5 4 3 2 1(.35)3 (.65)2
10(.35)3 (.65)2 .1811
3!(5 3)!
3 2 1 2 1
p(4)
5!(.35)4 (.65)5 4 5 4 3 2 1(.35)4 (.65)1
5(.35)4 (.65)1 .0488
4!(5 4)!
4 3 2 11
p(5)
5!(.35)5 (.65)5 5 5 4 3 2 1(.35)5 (.65)0
(.35)5 .0053
5!(5 5)!
5 4 3 2 11
The two properties of discrete random variables are that p(x) ≥ 0 for all x and Σp(x) = 1. From
above, all probabilities are greater than 0 and
Σp(x) = .1160 + .3124 + .3364 + .1811 + .0488 + .0053 = 1
4.24
d.
P(x ≥ 4) = p(4) + p(5) = .0488 + .0053 = .0541
a.
First, we must find the probability distribution of x. Define the following events:
C: {Chicken is contaminated}
N: {Chicken is not contaminated}
If 3 slaughtered chickens are randomly selected, then the possible outcomes are:
CCC, CCN, CNC, NCC, CNN, NCN, NNC, and NNN
Each of these outcomes are NOT equally likely since P(C) = 1/100 = .01. P(N) = 1 – P(C) = 1-−.01
= .99.
P(CCC) = P(C ∩ C ∩ C ) = P(C) P(C) P(C) = .01(.01)(.01) = .000001
P(CCN) = P(CNC) = P(NCC) = P(C ∩ C ∩ N ) = P(C) P(C) P(N) = .01(.01)(.99) = .000099
P(CNN) = P(NCN) = P(NNC) = P(C ∩ N ∩ N ) = P(C) P(N) P(N) = .01(.99)(.99) = .009801
P(NNN) = P(N ∩ N ∩ N ) = P(N) P(N) P(N) = .99(.99)(.99) = .970299.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
159
The variable x is defined as the number of contaminated chickens in the sample. The value of x for
each of the outcomes is:
Event
CCC
CCN
CNC
NCC
CNN
NCN
NNC
NNN
x
3
2
2
2
1
1
1
0
p(x)
.000001
.000099
.000099
.000099
.009801
.009801
.009801
.970299
The probability distribution of x is:
x
3
2
1
0
b.
p(x)
.000001
.000297
.029403
.970299
Using MINITAB, the probability graph for x is:
1.0
0.9
0.8
0.7
p(x)
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
x
4.25
c.
P(x ≤ 1) = P(x = 0) + P(x = 1) = .970299 + .029403 = .999702
a.
p(1) = (.23)(.77)1−1 = (.23)(.77)0 = .23. The probability that one would encounter a
contaminated cartridge on the first trial is .23.
b.
c.
p(5) = (.23)(.77)5−1 = (.23)(.77)4 = .0809. The probability that one would encounter a
the first contaminated cartridge on the fifth trial is .0809.
P(x ≥ 2) = 1 – P(x ≤ 1) = 1 – P(x = 1) = 1 − .23 = .77. The probability that the first contaminated
cartridge is found on the second trial or later is .77.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
160
4.26
Chapter 4
To find the probability distribution of x, we first list the possible values of x. For this exercise, the possible
values of x are −3, −1, and 5. Next, we list the number of cases, f(x), that result in the particular values of
x. To find the probability distribution of x, we divide the number of cases for each value of x, f(x), by the
total number of cases, 678. For x = −3, the probability is p(−3) = 68 / 678 = .100. For x = −1, the
probability is p(−1) = 71 / 678 = .105. For x = 5, the probability is p(5) = 539 / 678 = .795. The
probability distribution of x is:
x
−3
−1
5
Total
f(x)
68
71
539
678
p(x)
.100
.105
.795
1.000
Using MINITAB, the graph of the probability distribution is:
0.8
0.7
0.6
p(x)
0.5
0.4
0.3
0.2
0.1
0.0
-3
-2
-1
0
1
2
3
4
5
x
4.27
a.
20 100 20
20!
80!
20! 80!
80 79 78
1
0 3 - 0
0!(20 0)! 3!(80 3)! 0!20! 3!77!
3 2
p(0)
100!
100!
100 99 98
100
3
3!(100 3)!
3!97!
3 2
b.
82,160
.508
161, 700
20 100 20
20!
80!
20! 80!
80 79
20
1 3 - 1
1!(20 1)! 2!(80 2)! 1!19! 2!78!
2
p(1)
100!
100!
100 99 98
100
3
3!(100 3)!
3!97!
3 2
63, 200
.391
161, 700
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
c.
20 100 20
20!
80!
20! 80!
20 19
80
2 3 - 2
2!(20 2)! 1!(80 1)! 2!18! 1!79!
2
p(2)
100!
100!
100 99 98
100
3
3!(100 3)!
3!97!
3 2
d.
a.
15, 200
.094
161, 700
20 100 20
20!
80!
20!
20 19 18
1
1
3 3 - 0
3!(20 3)! 0!(80 0)! 3!17!
3 2
p(3)
100!
100!
100 99 98
100
3
3!(100 3)!
3!97!
3 2
4.28
161
E(x) =
1,140
.007
161, 700
xp( x)
All x
Firm A: E(x) = 0(.01) + 500(.01) + 1000(.01) + 1500(.02) + 2000(.35) + 2500(.30)
+ 3000(.25) + 3500(.02) + 4000(.01) + 4500(.01) + 5000(.01)
= 0 + 5 + 10 + 30 + 700 + 750 + 750 + 70 + 40 + 45 + 50
= 2450
Firm B: E(x) = 0(.00) + 200(.01) + 700(.02) + 1200(.02) + 1700(.15) + 2200(.30)
+ 2700(.30) + 3200(.15) + 3700(.02) + 4200(.02) + 4700(.01)
= 0 + 2 + 14 + 24 + 255 + 660 + 810 + 480 + 74 + 84 + 47
= 2450
b.
σ = σ2
σ2 =
(x μ)
2
p( x)
All x
Firm A: σ2 = (0 − 2450)2(.01) + (500 − 2450)2(.01) + ⋅⋅⋅ + (5000 − 2450)2(.01)
= 60,025 + 38,025 + 21,025 + 18,050 + 70,875 + 750 + 75,625
+ 22,050 + 24,025 + 42,025 + 65,025
= 437,500
σ = 661.44
Firm B: σ2 = (0 − 2450)2(.00) + (200 − 2450)2(.01) + ⋅⋅⋅ + (4700 − 2450)2(.01)
= 0 + 50,625 + 61,250 + 31,250 + 84,375 + 18,750 + 84,375
+ 31,250 + 61,250 + 50,625
= 492,500
σ = 701.78
Firm B faces greater risk of physical damage because it has a higher variance and standard deviation.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
162
4.29
Chapter 4
a.
The properties of valid probability distributions are:
p( x) = 1 and 0 ≤ p(x) ≤ 1 for all x.
For ARC a1:0 ≤ p(x) ≤ 1 for all x and
p( x) = .6 + .25 + .1 + .05 = 1.00
Thus, this is a valid probability distribution.
For ARC a2:0 ≤ p(x) ≤ 1 for all x and
p( x) = .6 + .3 + .1 = 1.00
Thus, this is a valid probability distribution.
For ARC a3:0 ≤ p(x) ≤ 1 for all x and
p( x) = .9 + .1 = 1.00
Thus, this is a valid probability distribution.
For ARC a4:0 ≤ p(x) ≤ 1 for all x and
p( x) = .9 + .1 = 1.00
Thus, this is a valid probability distribution.
For ARC a5:0 ≤ p(x) ≤ 1 for all x and
p( x) = .9 + .1 = 1.00
Thus, this is a valid probability distribution.
For ARC a6:0 ≤ p(x) ≤ 1 for all x and
p( x) = .7 + .25 + .05 = 1.00
Thus, this is a valid probability distribution.
b.
For Arc a1, P(x > 1) = P(x = 2) + P(x = 3) = .25 + .6 = .85
c.
For Arc a2, P(x > 1) = P(x = 2) = .6
For Arc a3, P(x > 1) = 0
For Arc a4, P(x > 1) = 0
For Arc a5, P(x > 1) = 0
For Arc a6, P(x > 1) = P(x = 2) = .7
d.
For Arc a1,
E ( x)
xp( x) 3(.60) 2(.25) 1(.10) 0(.05) 1.80 .50 .1 0 2.40
The average capacity of Arc a1 is 2.40.
For Arc a2,
E ( x)
xp( x) 2(.60) 1(.30) 0(.10) 1.20 .30 0 1.50
The average capacity of Arc a2 is 1.50.
For Arcs a3, a4, and a5,
E ( x)
xp( x) 1(.90) 0(.10) .90 0 .90
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
The average capacity of Arc a3 is 0.90. The average capacity of Arc a4 is 0.90.
The average capacity of Arc a5 is 0.90.
For Arc a6,
E ( x)
xp( x) 2(.70) 1(.25) 0(.10) 1.40 .25 0 1.65
The average capacity of Arc a6 is 1.65.
e.
For Arc a1,
σ 2 E ( x μ )
2
(x μ)
2
p( x)
(3 2.4) 2 (.60) (2 2.4) 2 (.25) (1 2.4) 2 (.10) (0 2.4) 2 (.05)
(.6) 2 (.60) ( .4) 2 (.25) (1.4)2 (.10) ( 2.4)2 (.05)
.216 .04 .196 .288 .74
σ .74 .86
We would expect most observations to fall within 2 standard deviations of the mean or
2.40 ± 2(.86) 2.40 ± 1.72 (.68, 4.12)
For Arc a2,
σ 2 E ( x μ )
2
(x μ)
2
p( x)
(2 1.5) (.60) (1 1.5) (.30) (0 1.5)2 (.10)
2
2
(.5)2 (.60) (.5)2 (.30) (1.5) 2 (.10)
.15 .075 .225 .45
σ .45 .67
We would expect most observations to fall within 2 standard deviations of the mean or
1.50 ± 2(.67) 1.50 ± 1.34 (.16, 2.84)
For Arcs a3, a4, and a5,
σ 2 E ( x μ )
2
(x μ)
2
p( x)
(1 .9)2 (.90) (0 .9)2 (.10)
(.1)2 (.90) (.9)2 (.10)
.009 .081 .090
σ .09 .30
We would expect most observations to fall within 2 standard deviations of the mean or
.90 ± 2(.30) .90 ± .60 (.30, 1.50)
For Arc a6,
σ 2 E ( x μ )
2
(x μ)
2
p( x)
(2 1.65)2 (.70) (1 1.65)2 (.25) (0 1.65)2 (.05)
(.35)2 (.70) (.65) 2 (.25) (1.65) 2 (.05)
.08575 .105625 .136125 .3275
σ .3275 .57
We would expect most observations to fall within 2 standard deviations of the mean or
1.65 ± 2(.57) 1.65 ± 1.14 (.51, 2.79)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
163
164
4.30
Chapter 4
a.
If a large number of measurements are observed, then the relative frequencies should be very good
estimators of the probabilities.
b.
E(x) =
xp( x) = 1(.01) + 2(.04) + 3(.04) + 4(.08) + 5(.10) + 6(.15) + 7(.25) + 8(.20)
+ 9(.08) + 10(.05)
= .01 + .08 + .12 + .32 + .50 + .90 + 1.75 + 1.60 + .72 + .50
= 6.50
The average number of checkout lanes per store is 6.5.
c.
σ2 =
(x μ)
2
p( x) = (1 − 6.5)2(.01) + (2 − 6.5)2(.04) + (3 − 6.5)2(.04)
All x
+ (4 − 6.5)2(.08) + (5 − 6.5)2(.10) + (6 − 6.5)2(.15)
+ (7 − 6.5)2(.25) + (8 − 6.5)2(.20) + (9 − 6.5)2(.08)
+ (10 − 6.5)2(.05)
= .3025 + .8100 + .4900 + .5000 + .2250 + .0375 + .0625
+ .4500 + .5000 + .6125
= 3.99
σ=
d.
3.99 = 1.9975
Chebyshev's Rule says that at least 0 of the observations should fall in the interval μ ± σ.
Chebyshev's Rule says that at least 75% of the observations should fall in the interval
μ ± 2σ.
e.
μ ± σ 6.5 ± 1.9975 (4.5025, 8.4975)
P(4.5025 ≤ x ≤ 8.4975) = .10 + .15 + .25 + .20 = .70
This is at least 0.
μ ± 2σ 6.5 ± 2(1.9975) 6.5 ± 3.995 (2.505, 10,495)
P(2.505 ≤ x ≤ 10.495) = .04 + .08 + .10 + .15 + .25 + .20 + .08 + .05 = .95
This is at least .75 or 75%.
4.31
a.
Let x = the potential flood damages. Since we are assuming if it rains the business will incur
damages and if it does not rain the business will not incur any damages, the probability distribution
of x is:
x
p(x)
b.
0
.7
300,000
.3
The expected loss due to flood damage is
E(x) =
xp(x) = 0(.7) + 300,000(.3) = 0 + 90,000 = $90,000
All x
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Random Variables and Probability Distributions
4.32
165
Let x = winnings in the Florida lottery. The probability distribution for x is:
x
p(x)
−$1
22,999,999/23,000,000
$6,999,999
1/23,000,000
The expected net winnings would be:
μ = E(x) = (−1)(22,999,999/23,000,000) + 6,999,999(1/23,000,000) = −$.70
The average winnings of all those who play the lottery is −$.70.
4.33
a.
Since there are 20 possible outcomes that are all equally likely, the probability of any of the 20
numbers is 1/20. The probability distribution of x is:
P(x = 5) = 1/20 = .05; P(x = 10) = 1/20 = .05; etc.
x
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
p(x) .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05 .05
b.
E(x) =
xp( x) = 5(.05) + 10(.05) + 15(.05) + 20(.05) + 25(.05) + 30(.05) + 35(.05)
+ 40(.05) + 45(.05) + 50(.05) + 55(.05) + 60(.05) + 65(.05) + 70(.05) + 75(.05)
+ 80(.05) + 85(.05) + 90(.05) + 95(.05) + 100(.05) = 52.5
c.
σ2 = E(x − μ)2 =
( x μ ) p( x) = (5 − 52.5) (.05) + (10 - 52.5) (.05)
2
2
2
+ (15 − 52.5)2(.05) + (20 − 52.5)2(.05) + (25 − 52.5)2(.05) + (30 − 52.5)2(.05)
+ (35 − 52.5)2(.05) + (40 − 52.5)2(.05) + (45 − 52.5)2(.05) + (50 − 52.5)2(.05)
+ (55 − 52.5)2(.05) + (60 − 52.5)2(.05) + (65 − 52.5)2(.05) + (70 − 52.5)2(.05)
+ (75 − 52.5)2(.05) + (80 − 52.5)2(.05) + (85 − 52.5)2(.05) + (90 − 52.5)2(.05)
+ (95 − 52.5)2(.05) + (100 − 52.5)2(.05)
= 831.25
σ = σ 2 831.25 = 28.83
Since the uniform distribution is not mound-shaped, we will use Chebyshev's theorem to describe the
data. We know that at least 8/9 of the observations will fall with 3 standard deviations of the mean
and at least 3/4 of the observations will fall within 2 standard deviations of the mean. For this
problem,
μ ± 2σ 52.5 ± 2(28.83) 52.5 ± 57.66 (−5.16, 110.16). Thus, at least 3/4 of the data will fall
between −5.16 and 110.16. For our problem, all of the observations will fall within 2 standard
deviations of the mean. Thus, x is just as likely to fall within any interval of equal length.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
166
Chapter 4
d.
If a player spins the wheel twice, the total number of outcomes will be 20(20) = 400. The sample
space is:
5, 5
10, 5
5,10 10,10
5,15 10,15
.
.
.
.
.
.
5,100 10,100
15, 5
15,10
15,15
.
.
.
15,100
20, 5
20,10
20,15
.
.
.
20,100
25, 5...
100, 5
25,10... 100,10
25,15... 100,15
.
.
.
.
.
.
25,100... 100,100
Each of these outcomes are equally likely, so each has a probability of 1/400 = .0025.
Now, let x equal the sum of the two numbers in each sample. There is one sample with a sum of 10,
two samples with a sum of 15, three samples with a sum of 20, etc. If the sum of the two numbers
exceeds 100, then x is zero. The probability distribution of x is:
e.
x
p(x)
0
.5250
10
.0025
15
.0050
20
.0075
25
.0100
30
.0125
35
.0150
40
.0175
45
.0200
50
.0225
55
.0250
60
.0275
65
.0300
70
.0325
75
.0350
80
.0375
85
.0400
90
.0425
95
.0450
100
.0475
We assumed that the wheel is fair, or that all outcomes are equally likely.
f.
μ = E(x) =
xp( x) = 0(.5250) + 10(.0025) + 15(.0050) + 20(.0075) + ...+ 100(.0475)
= 33.25
σ2 = E(x − μ)2 =
(x - μ)
2
p( x) = (0 − 33.25)2(.525) + (10 − 33.25)2(.0025)
+ (15 − 33.25)2(.0050) + (20 − 33.25)2(.0075) + ...+ (100 − 33.25)2(.0475)
= 1471.3125
σ = σ 2 1471.3125 = 38.3577
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Random Variables and Probability Distributions
167
g.
P(x = 0) = .525
h.
Given that the player obtains a 20 on the first spin, the possible values for x (sum of the two spins)
are 0 (player spins 85, 90, 95, or 100 on the second spin), 25, 30, ..., 100. In order to get an x of 25,
the player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on the
second spin order to get an x of 30, etc. Since all of the outcomes are equally likely on the second
spin, the distribution of x is:
x
p(x)
0
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
.20
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
.05
i.
The probability that the players total score will exceed one dollar is the probability that x is zero.
P(x = 0) = .20
j.
Given that the player obtains a 65 on the first spin, the possible values for x (sum of the two spins)
are 0 (player spins 40, 45, 50, up to 100 on second spin), 70, 75, 80,..., 100. In order to get an x of
70, the player would spin a 5 on the second spin. Similarly, the player would have to spin a 10 on
the second spin in order to get an x of 75, etc. Since all of the outcomes are equally likely on the
second spin, the distribution of x is:
x
p(x)
0
70
75
80
85
90
95
100
.65
.05
.05
.05
.05
.05
.05
.05
The probability that the players total score will exceed one dollar is the probability that x is zero.
P(x = 0) = .65.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
168
4.34
Chapter 4
a.
Each point in the system can have one of 2 status levels, “free” or “obstacle”. Define the following
events:
AF: {Point A is free}
BF: {Point B is free}
CF: {Point C is free}
AO: {Point A is obstacle}
BO: {Point B is obstacle}
CO: {Point C is obstacle}
Thus, the sample points for the space are:
AFBFCF, AFBFCO, AFBOCF, AFBOCO, AOBFCF, AOBFCO, AOBOCF, AOBOCO
b.
Since it is stated that the probability of any point in the system having a “free” status is .5, the
probability of any point having an “obstacle” status is also .5, Thus, the probability of each of the
sample points above is P(AiBiCi) = .5(.5)(.5) = .125.
The values of Y, the number of free links in the system, for each sample point are listed below. A
link is free if both the points are free. Thus, a link from A to B is free if A is free and B is free. A
link from B to C is free if B is free and C is free.
Sample point
Y
Probability
AFBFCF
2
.125
AFBFCO
1
.125
AFBOCF
0
.125
AFBOCO
0
.125
AOBFCF
1
.125
AOBFCO
0
.125
AOBOCF
0
.125
AOBOCO
0
.125
The probability distribution for Y is:
Y
Probability
0
.625
1
.250
2
.125
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Random Variables and Probability Distributions
4.35
169
Let x = bookie's earnings per dollar wagered. Then x can take on values $1 (you lose) and $-5 (you win).
The only way you win is if you pick 3 winners in 3 games. If the probability of picking 1 winner in 1
game is .5, then P(www) = p(w)p(w)p(w) = .5(.5)(.5) = .125 (assuming games are independent).
Thus, the probability distribution for x is:
x
p(x)
$1 .875
$-5 .125
4.36
E(x) =
xp( x) = 1(.875) − 5(.125) = .875 − .625 = $.25
a.
6!
6!
6 5 4 3 2 1
15
2!(6 2)! 2!4! (2 1)(4 3 2 1)
b.
5
5!
5!
5 4 3 2 1
2 2!(5 2)! 2!3! (2 1)(3 2 1) 10
c.
7
7!
7!
7 6 5 4 3 2 1
0 0!(7 0)! 0!7! (1)(7 6 5 4 3 2 1) 1
(Note: 0! = 1)
4.37
d.
6
6!
6!
6 5 4 3 2 1
6 6!(6 6)! 6!0! (6 5 4 3 2 1)(1) 1
e.
4
4!
4!
4 3 2 1
3 3!(4 3)! 3!1! (3 2 1)(1) 4
a.
x is discrete. It can take on only six values.
b.
This is a binomial distribution.
c.
5
5!
5 4 3 2 1
(.7)0(.3)5 =
(1)(.00243) = .00243
p(0) = (.7)0(.3)5-0 =
0
0!5!
1
5 43 21
5
5!
(.7)1(.3)4 = .02835
p(1) = (.7)1(.3)5-1 =
1!4!
1
5
5!
(.7)2(.3)3 = .1323
p(2) = (.7)2(.3)5-2 =
2!3!
2
5
5!
(.7)3(.3)2 = .3087
p(3) = (.7)3(.3)5-3 =
3!2!
3
5
5!
(.7)4(.3)1 = .36015
p(4) = (.7)4(.3)5-4 =
4!1!
4
5
5!
(.7)5(.3)0 = .16807
p(5) = (.7)5(.3)5-5 =
5!0!
5
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
170
4.38
Chapter 4
d.
μ = np = 5(.7) = 3.5
σ = npq 5(.7)(.3) 1.0247
e.
μ ± 2σ = 3.5 ± 2(1.0247) (1.4506, 5.5494)
a.
3
3!
3 2 1
(.3)0(.7)3 =
(1)(.7)3 = .343
p(0) = (.3)0(.7)3-0 =
0!3!
1 3 2 1
0
3
3!
(.3)1(.7)2 = .441
p(1) = (.3)1(.7)3-1 =
1!2!
1
3
5!
(.3)2(.7)1 = .189
p(2) = (.3)2(.7)3-2 =
2!1!
2
3
5!
(.3)3(.7)0 = .027
p(3) = (.3)3(.7)3-3 =
3!0!
3
4.39
x
p(x)
0
1
2
3
.343
.441
.189
.027
a.
P(x = 1) =
5!
5 4 3 2 1
(.2)1(.8) 4 =
(.2)1(.8) 4 = 5(.2)1(.8)4 = .4096
1!4!
(1)(4 3 2 1)
b.
P(x = 2) =
4!
4 3 2 1
(.6) 2(.4) 2 =
(.6) 2(.4) 2 = 6(.6)2(.4)2 = .3456
2!2!
(2 1)(2 1)
c.
P(x = 0) =
3!
3 2 1
(.7) 0(.3) 3 =
(.7) 0(.3) 3 = 1(.7)0(.3)3 = .027
0!3!
(1)(3 2 1)
d.
P(x = 3) =
5!
5 4 3 2 1
(.1) 3(.9) 2 =
(.1) 3(.9) 2 = 10(.1)3(.9)2 = .0081
3!2!
(3 2 1)(2 1)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.40
4.41
e.
P(x = 2) =
4!
4 3 2 1
(.4) 2(.6) 2 =
(.4) 2(.6) 2 = 6(.4)2(.6)2 = .3456
2!2!
(2 1)(2 1)
f.
P(x = 1) =
3!
3 2 1
(.9)1(.1) 2 =
(.9)1(.1) 2 = 3(.9)1(.1)2 = .027
1!2!
(1)(2 1)
a.
P(x = 2) = P(x ≤ 2) − P(x ≤ 1) = .167 − .046 = .121 (from Table II, Appendix B)
b.
P(x ≤ 5) = .034
c.
P(x > 1) = 1 − P(x ≤ 1) = 1 − .919 = .081
d.
P(x < 10) = P(x ≤ 9) = 0
e.
P(x ≥ 10) = 1 − P(x ≤ 9) = 1 − .002 = .998
f.
P(x = 2) = P(x ≤ 2) − P(x ≤ 1) = .206 − .069 = .137
a.
μ = np = 25(.5) = 12.5
σ2 = np(1 − p) = 25(.5)(.5) = 6.25
σ = σ 2 6.25 2.5
b.
μ = np = 80(.2) = 16
σ2 = np(1 − p) = 80(.2)(.8) = 12.8
σ = σ 2 12.8 3.578
c.
μ = np = 100(.6) = 60
σ2 = np(1 − p) = 100(.6)(.4) = 24
σ = σ 2 24 4.899
d.
μ = np = 70(.9) = 63
σ2 = np(1 − p) = 70(.9)(.1) = 6.3
σ = σ 2 6.3 2.510
e.
f.
μ = np = 60(.8) = 48
σ2 = np(1 − p) = 60(.8)(.2) = 9.6
σ = σ 2 9.6 3.098
μ = np = 1,000(.04) = 40
σ2 = np(1 − p) = 1,000(.04)(.96) = 38.4
σ = σ 2 38.4 6.197
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171
172
4.42
Chapter 4
x is a binomial random variable with n = 4.
a.
If the probability distribution of x is symmetric, p(0) = p(4) and p(1) = p(3).
n
Since p(x) = pxqn-x x = 0, 1, ... , n,
x
When n = 4,
4 0 4 4 4 0
4! 0 4 4! 4 0
4
4
0 p q 4 p q 0!4! p q 4!0! p q q p p q
Since p + q = 1, p = .5
Therefore, the probability distribution of x is symmetric when p = .5.
b.
If the probability distribution of x is skewed to the right, then the mean is greater than the median.
Therefore, there are more small values in the distribution (0, 1) than large values (3, 4). Therefore, p
must be smaller than .5. Let p = .2 and the probability distribution of x will be skewed to the right.
c.
If the probability distribution of x is skewed to the left, then the mean is smaller than the median.
Therefore, there are more large values in the distribution (3, 4) than small values (0, 1). Therefore, p
must be larger than .5. Let p = .8 and the probability distribution of x will be skewed to the left.
d.
In part a, x is a binomial random variable with n = 4 and p = .5.
4
p(x) = .5x.54-x
x
x = 0, 1, 2, 3, 4
4
4! 4
.5 = 1(.5)4 = .0625
p(0) = .50.5 4
0!4!
0
4
4! 4
.5 = 4(.5)4 = .25
p(1) = .51.53
1!3!
1
4
4! 4
.5 = 6(.5)4 = .375
p(2) = .5 2.5 2
2!2!
2
p(3) = p(1) = .25 (since the distribution is symmetric)
p(4) = p(0) = .0625
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Random Variables and Probability Distributions
173
The probability distribution of x in tabular form is:
x
0
1
2
3
4
p(x)
.0625
.25
.375
.25
.0625
μ = np = 4(.5) = 2
The graph of the probability distribution of x when
n = 4 and p = .5 is as follows.
In part b, x is a binomial random variable with
n = 4 and p = .2.
4
p(x) = .2 x.8 4 x
x
x = 0, 1, 2, 3, 4
4
p(0) = .20.84 = 1(1).84 = .4096
0
4
p(1) = .21.83 = 4(.2)(.8)3 = .4096
1
4
p(2) = .22.82 = 6(.2)2(.8)2 = .1536
2
4
p(3) = .23.81 = 4(.2)3(.8) = .0256
3
4
p(4) = .24.80 = 1(.2)4(1) = .0016
4
The probability distribution of x in tabular form is:
x
0
1
2
3
4
p(x)
.4096
.4096
.1536
.0256
.0016
μ = np = 4(.2) = .8
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174
Chapter 4
The graph of the probability distribution of x
when n = 4 and p = .2 is as follows:
In part c, x is a binomial random variable with n = 4 and p = .8.
4
p(x) = .8 x.2 4- x
x
x = 0, 1, 2, 3, 4
4
p(0) = .80.24 = 1(1).24 = .0016
0
4
p(1) = .81.23 = 4(.8)(.2)3 = .0256
1
4
p(2) = .82.22 = 6(.8)2(.2)2 = .1536
2
4
p(3) = .83.21 = 4(.8)3(.2) = .4096
3
4
p(4) = .84.20 = 1(.8)4(1) = .4096
4
The probability distribution of x in tabular form is:
x
0
1
2
3
4
p(x)
.0016
.0256
.1536
.4096
.4096
Note: The distribution of x when n = 4 and p = .2 is the reverse of the distribution of
x when n = 4 and p = .8.
μ = np = 4(.8) = 3.2
The graph of the probability distribution of x when
n = 4 and p = .8 is as follows:
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Random Variables and Probability Distributions
4.43
175
e.
In general, when p = .5, a binomial distribution will be symmetric regardless of the value of n. When
p is less than .5, the binomial distribution will be skewed to the right; and when p is greater than .5, it
will be skewed to the left. (Refer to parts a, b, and c.)
a.
We will check the 5 characteristics of a binomial random variable.
1.
2.
3.
4.
5.
The experiment consists of 100 identical trials.
There are only 2 possible outcomes for each trial. Let S = internet user goes online at
home using a wireless network and F = internet user goes online at home without using a
wireless network.
The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .20 and
q = 1 – p = 1 .20 = .80.
The trials are independent.
The binomial random variable x is the number of internet users in 100 trials who go online at
home using a wireless connection.
Thus, x is a binomial random variable.
4.44
b.
From the exercise, p = .20. For any internet user who goes online at home, the probability of using a
wireless connection is .20.
c.
µ = E(x) = np = 100(.20) = 20. On average, for every 100 internet users who go online at home, 20
will use a wireless connection.
a.
We will check the 5 characteristics of a binomial random variable.
1.
2.
3.
4.
5.
The experiment consists of n = 5 identical trials. We have to assume that the number of
bottled water brands is large.
There are only 2 possible outcomes for each trial. Let S = brand of bottled water used tap
water and F = brand of bottled water did not use tap water.
The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .25 and
q = 1 – p = 1 - .25 = .75.
The trials are independent.
The binomial random variable x is the number of brands in the 5 trials that used tap water.
If the total number of brands of bottled water is large, then the above characteristics will be basically
true. Thus, x is a binomial random variable.
b.
5
The formula for the probability distribution for x is p( x) .25x (.75)5 x ,
x
for x = 1, 2, 3, 4, 5.
c.
5
5!
.252.753 .2637
P( x 2) .252 (.75)5 2
2!3!
2
d.
5
5
P( x 1) P ( x 0) P( x 1) .250 (.75)5 0 .251 (.75)51
0
1
5!
5!
.250.755
.251.754 .2373 .3955 .6328
0!5!
1!4!
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176
4.45
Chapter 4
a.
Let x = number of small businesses owned by non-Hispanic whites that are female
owned in 200 trials. Then x is a binomial random variable with n = 200 and p = .27.
μ E ( x) np 200(.27) 54
b.
Let x = number of small businesses owned by non-Hispanic whites that are female
owned in 8 trials. Then x is a binomial random variable with n = 8 and p = .27.
n
8
P( x 0) p x (1 p)n x .270 (.73)8 0 .738 .0806
x
0
n
8
8!
P( x 4) p x (1 p) n x .27 4 (.73)8 4
.27 4.734 .1056
4!(8 4)!
x
4
4.46
a.
We will check the 5 characteristics of a binomial random variable.
1.
2.
The experiment consists of n identical trials.
There are only 2 possible outcomes for each trial. Let S = general practice physician in
the United States does not recommend medicine as a career and F = general practice physician
in the United States does recommend medicine as a career.
The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .60 and
q = 1 – p = 1 - .60 = .40.
The trials are independent.
The binomial random variable x is the number of general practice physicians in the United
States in n trials who do not recommend medicine as a career.
3.
4.
5.
Thus, x is a binomial random variable.
b.
From the information given, p = .60.
c.
µ = E(x) = np = 25(.60) = 15
σ npq 25(.60)(.40) 6 2.4495
4.47
d.
From Table II, Appendix B, with n = 25 and p = .60, P(x ≥ 1) = 1 – P(x = 0) = 1 – .000 = 1.000.
a.
We will check the 5 characteristics of a binomial random variable.
1.
2.
3.
4.
5.
The experiment consists of n = 20 identical trials.
There are only 2 possible outcomes for each trial. Let S = intruding object is detected and
F = intruding object is not detected.
The probability of success (S) is the same from trial to trial. For each trial, p =
P(S) =
.8 and q = 1 – p = 1 − .8 = .2.
The trials are independent.
The binomial random variable x is the number of intruding objects in the 20 trials that are
detected.
Thus, x is a binomial random variable.
b.
For this experiment, n = 20 and p = .8.
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Random Variables and Probability Distributions
c.
177
Using Table II, Appendix B, with n = 20 and p = .8,
P(x = 15) = P(x ≤ 15) − P(x ≤ 14) = .370 − .196 = .174
d.
Using Table II, Appendix B, with n = 20 and p = .8,
P(x ≥ 15) = 1 − P(x ≤ 14) = 1 − .196 = .804
4.48
e.
E(x) = np = 20(.8) = 16. For every 20 intruding objects, SBIRS will detect an average of 16.
a.
Let x = number of commissioners out of 4 who vote in favor of an issue. Then x is a binomial
random variable with n = 4 and p = .5 (since they are equally likely to vote for or against an issue).
The probability that your vote counts is equal to P(x = 2).
P( x 2)
b.
Let x = number of commissioners out of 2 who vote in favor of an issue. Then x is a binomial
random variable with n = 2 and p = .5 (since they are equally likely to vote for or against an issue).
The probability that your vote counts is equal to P(x = 1).
P( x 1)
4.49
4!
4 3 2 1 2
.52 (.5) 4 2
.5 (.5) 2 .375
2!(4 2)!
2 1 2 1
2!
2 1 1 1
.51 (.5)2 1
.5 (.5) .5
1!(2 1)!
1 1
Let x = number of major bridges in Denver that will have a rating of 4 or below in 2020 in 10 trials. Then
x has an approximate binomial distribution with n = 10 and p = .09.
a.
P( x 3) 1 P( x 2) 1 P( x 0) P ( x 1) P( x 2)
10
10
10
1 .09 0 (.91)10 0 .091 (.91)10 1 .092 (.91)10 2
0
1
2
1
b.
4.50
10!
10! 1 9 10!
.090.9110
.09 .91
.092.918 1 .389 .385 .171 .055
0!10!
1!9!
2!8!
Since the probability of seeing at least 3 bridges out of 10 with ratings of 4 or less is so small, we can
conclude that the forecast of 9% of all major Denver bridges will have ratings of 4 or less in 2020 is
too small. There would probably be more than 9%.
Let x = number of packets observed by a network sensor in 150 trials. Then x has an approximate binomial
distribution with n = 150 and p = .001.
The virus will be detected if at least 1 packets is observed.
150
150!
P( x 1) 1 P ( x 0) 1
.0010 (.999)150 0 1
.999150 1 .8606 .1394
0!150!
0
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178
4.51
Chapter 4
Define the following events:
A: {Taxpayer is audited}
B: {Taxpayer has income less than $1 million)
C: {Taxpayer has income of $1 million or higher}
a.
From the information given in the problem,
P(A│B) = 1/100 = .01
P(A│C) = 9/100 = .09
b.
Let x = number of taxpayers with incomes under $1 million who are audited. Then x is a binomial
random variable with n = 5 and p = .01.
5
5!
.011 .99 (4) .0480
P(x = 1) = .011 .99 (51)
1!4!
1
P(x > 1) = 1 − [P(x = 0) + P(x = 1)]
5
= 1 .010 .99 (5 0) .0480
0
5!
= 1
.010 .99 5 .0480
0!5!
= 1 − [.9510 + .0480] = 1 − .9990 = .0010
c.
Let x = number of taxpayers with incomes of $1 million or more who are audited. Then x is a
binomial random variable with n = 5 and p = .09.
5
5!
.091 .914 = .3086
P(x = 1) = .091 .91(51)
1!4!
1
P(x > 1) = 1 − [P(x = 0) + P(x = 1)]
5
= 1 .09 0 .91(5 0) .3086
0
5!
.09 0 .915 .3086
= 1
0!5!
= 1 − [.6240 + .3086] = 1 − .9326= .0674
d.
Let x = number of taxpayers with incomes under $1 million who are audited. Then x is a binomial
random variable with n = 2 and p = .01.
Let y = number of taxpayers with incomes $1 million or more who are audited. Then y is a binomial
random variable with n = 2 and p = .09.
2
2!
.010 .99 2 = .9801
P(x = 0) = .010 .99 (2 0)
0!2!
0
2
2!
.09 0 .912 = .8281
P(y = 0) = .09 0 .91(2 0)
0!2!
0
P(x = 0)P(y = 0) = .9801(.8281) = .8116
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Random Variables and Probability Distributions
e.
4.52
179
We must assume that the variables defined as x and y are binomial random variables. We must
assume that the trials are identical, the probability of success is the same from trial to trial, and that
the trials are independent.
Let x = number of UWG male students out of 50 who have gambled on sports in the past year.
Then x will have an approximate binomial distribution with n = 50 and p = .60.
µ = E(x) = np = 50(.60) = 30.
σ npq 50(.60)(.40) 12 3.464
We would expect that most of the observations would fall within 2 standard deviations of the mean. Thus,
a likely range for the number of male students who have gambled on sports in the last year would be:
30 ± 2(3.464) 30 ± 6.928 (23.072, 36.928).
The range from 24 to 36 would likely include the actual number of male students in a sample of 50 who
gambled on sports in the last year.
4.53
a.
μ E ( x) np 800(.70) 560
σ npq 800(.70)(.30) 168 12.96
b.
Half of the 800 food items would be 400. A value of x = 400 would have a z-score of:
z
xμ
σ
400 560
12.35
12.96
Since the z-score associated with 400 items is so small (−12.35), it would be virtually impossible to
observe less than half with any pesticides if the 70% value was correct.
4.54
Assuming the supplier's claim is true,
μ = np = 500(.001) = .5
σ = npq 500(.001)(.999) .4995 .707
If the supplier's claim is true, we would only expect to find .5 defective switches in a sample of size 500.
Therefore, it is not likely we would find 4.
Based on the sample, the guarantee is probably inaccurate.
Note: z
xμ
σ
4 .5
4.95
.707
This is an unusually large z-score.
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180
4.55
Chapter 4
a.
We must assume that the probability that a specific type of ball meets the requirements is always the
same from trial to trial and the trials are independent. To use the binomial probability distribution,
we need to know the probability that a specific type of golf ball meets the requirements.
b.
For a binomial distribution,
μ = np
σ = npq
In this example, n = two dozen = 2 ⋅ 12 = 24.
p = .10
(Success here means the golf ball does not meet standards.)
q = .90
μ = np = 24(.10) = 2.4
σ = npq 24(.10)(.90) = 1.47
c.
In this situation,
p = Probability of success
= Probability golf ball does meet standards
= .90
q = 1 − .90 = .10
n = 24
E(x) = μ = np = 24(.90) = 21.60
σ = npq 24(.10)(.90) = 1.47 (Note that this is the same as in part b.)
4.56
a.
For this test, n = 20 and p = .10. Then x is a binomial random variable with n = 20 and p = .10.
Using Table II, Appendix, with n = 20 and p = .10,
P(x ≤ 1) = .392
b.
For the experiment in part a, the level of confidence is 1 − P(x ≤ 1) = 1 − .392 = .608. Since this
value is not close to 1, this would not be an acceptable level.
c.
Suppose we increased n from 20 to 25. Using Table II, Appendix B, with n = 25 and p = .10,
P(x ≤ 1) = .271. This value is smaller than the value found in part a.
Now, suppose we keep n = 20, but change K to 0 instead of 1. Using Table II, Appendix B, with
n = 20 and p = .10,
P(x ≤ 0) = .122. This value is again, smaller than the value found in part a.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
d.
181
Suppose we let K = 0. Now, we need to find n such that the level of confidence ≥ .95, which means
that P(x = 0) ≤ .05.
n
P( x 0) .10 (.9) n 0 .05
0
n! n
.9 .05
0!n!
.9n .05
ln(.9n ) ln(.05)
nln(.9) ln(.05)
ln(.05) 2.99573
n
28.4
.10536
ln(.9)
Thus, if K = 0, then we need a sample size of 28 to get a level of confidence of at least .95.
Now, suppose K = 1. Now, we need to find n such that the level of confidence is at least .95, which
means that P(x ≤ 1) ≤ .05.
n
n
P( x 1) P( x 0) P( x 1) .10 (.9) n 0 .11 (.9) n 1 .05
0
1
n! n
n!
.9
.11.9n 1 .05
0!n!
1!(n 1)!
.9n n.11.9n 1 .05
.9n 1 (.9 .1n) ln(.05)
From here, we will use trial and error.
For n = 30, .930-1(.9+.1(30)) = .1837
n
.9n-1(.9+.1n)
30
.930-1(.9+.1(30)) = .1837
40
.940-1(.9+.1(40)) = .0805
45
.945-1(.9+.1(45)) = .0524
46
.946-1(.9+.1(46)) = .0480
Thus, for K = 1, we would need a sample size of 46 to get a level of confidence of at least .95.
4.57
a.
The random variable x is discrete since it can assume a countable number of values (0, 1, 2, ...).
b.
This is a Poisson probability distribution with λ = 3.
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182
Chapter 4
c.
In order to graph the probability distribution, we need to know the probabilities for the possible
values of x. Using Table III of Appendix B with λ = 3:
p(0) = .050
p(1) = P(x ≤ 1) − P(x = 0) = .199 − .050 = .149
p(2) = P(x ≤ 2) − P(x ≤ 1) = .423 − .199 = .224
p(3) = P(x ≤ 3) − P(x ≤ 2) = .647 − .423 = .224
p(4) = P(x ≤ 4) − P(x ≤ 3) = .815 − .647 = .168
p(5) = P(x ≤ 5) − P(x ≤ 4) = .916 − .815 = .101
p(6) = P(x ≤ 6) − P(x ≤ 5) = .966 − .916 = .050
p(7) = P(x ≤ 7) − P(x ≤ 6) = .988 − .966 = .022
p(8) = P(x ≤ 8) − P(x ≤ 7) = .996 − .988 = .008
p(9) = P(x ≤ 9) − P(x ≤ 8) = .999 − .996 = .003
p(10) ≈ .001
The probability
distribution of x in
graphical form is:
d.
μ=λ=3
σ2 = λ = 3
σ = 3 = 1.73
e.
The mean of x is the same as the mean of the probability distribution, μ = λ = 3.
The standard deviation of x is the same as the standard deviation of the probability distribution, σ =
1.7321.
4.58
μ = λ = 1.5
Using Table III of Appendix B:
a.
P(x ≤ 3) = .934
b.
P(x ≥ 3) = 1 − P(x ≤ 2) = 1 − .809 = .191
c.
P(x = 3) = P(x ≤ 3) − P(x ≤ 2) = .934 − .809 = .125
d.
P(x = 0) = .223
e.
P(x > 0) = 1 − P(x = 0) = 1 − .223 = .777
f.
P(x > 6) = 1 − P(x ≤ 6) = 1 − .999 = .001
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.59
4.60
a.
r N r
x n x
P(x = 1) =
N
n
3 5 3
3! 2!
1 3 1
3(1)
1!2! 2!0!
= .3
5!
10
5
3
3!2!
b.
r N r 3 9 3
3! 6!
x n x 3 5 3
3!0!
2!4! 1(15) = .119
P(x = 3) =
9!
N
9
126
n
5
5!4!
c.
r N r 2 4 2
2! 2!
x n x 2 2 2
1(1)
2!0! 0!2!
= .167
P(x = 2) =
4!
6
N
4
n
2
2!2!
d.
r N r 2 4 2
2! 2!
x n x 0 2 0
0!2!
2!0! 1(1) = .167
P(x = 0) =
4!
6
N
4
n
2
2!2!
For N = 8, n = 3, and r = 5,
a.
r N r 5 8 5
5! 3!
x n x 1 3 1
5(3)
P(x = 1) =
1!4! 2!1!
= .268
8!
56
N
8
n
3
3!5!
b.
r N r 5 8 5
5! 3!
x n x 0 3 0
0!5!
3!0! 1(1) = .018
P(x = 0) =
8!
56
N
8
n
3
3!5!
c.
r N r 5 8 5
5! 3!
x n x 3 3 3
10(1)
3!2! 0!3!
= .179
P(x = 3) =
8!
56
N
8
n
3
3!5!
d.
P(x ≥ 4) = P(x = 4) + P(x = 5) = 0
Since the sample size is only 3, there is no way to get 4 or more successes in only 3 trials.
4.61
a.
For λ = 1, P(x ≤ 2) = .920 (from Table III, Appendix B)
b.
For λ = 2, P(x ≤ 2) = .677
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183
184
4.62
Chapter 4
c.
For λ = 3, P(x ≤ 2) = .423
d.
The probability decreases as λ increases. This is reasonable because λ is equal to the mean. As the
mean increases, the probability that x is less than a particular value will decrease.
a.
To graph the Poisson probability distribution with λ = 5, we need to calculate p(x) for x = 0 to 15.
Using Table III, Appendix B,
p(0) = .007
p(1) = P(x ≤ 1) − P(x ≤ 0) = .040 − .007 = .033
p(2) = P(x ≤ 2) − P(x ≤ 1) = .125 − .040 = .085
p(3) = P(x ≤ 3) − P(x ≤ 2) = .265 − .125 = .140
p(4) = P(x ≤ 4) − P(x ≤ 3) = .440 − .265 = .175
p(5) = P(x ≤ 5) − P(x ≤ 4) = .616 − .440 = .176
p(6) = P(x ≤ 6) − P(x ≤ 5) = .762 − .616 = .146
p(7) = P(x ≤ 7) − P(x ≤ 6) = .867 − .762 = .105
p(8) = P(x ≤ 8) − P(x ≤ 7) = .932 − .867 = .065
p(9) = P(x ≤ 9) − P(x ≤ 8) = .968 − .932 = .036
p(10) = P(x ≤ 10) − P(x ≤ 9) = .986 − .968 = .018
p(11) = P(x ≤ 11) − P(x ≤ 10) = .995 − .986 = .009
p(12) = P(x ≤ 12) − P(x ≤ 11) = .998 − .995 = .003
p(13) = P(x ≤ 13) − P(x ≤ 12) = .999 − .998 = .001
p(14) = P(x ≤ 14) − P(x ≤ 13) = 1.000 − .999 = .001
p(15) = P(x ≤ 15) − P(x ≤ 14) = 1.000 − 1.000 = .000
The graph is shown at right:
b.
μ=λ=5
σ = λ = 5 = 2.2361
μ ± 2σ 5 ± 2(2.2361) 5 ± 4.4722 (.5278, 9.4722)
c.
P(.5278 < x < 9.4722) = P(1 ≤ x ≤ 9) = P(x ≤ 9) − P(x = 0)
= .968 − .007 = .961
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Random Variables and Probability Distributions
4.63
4.64
185
For this problem, N = 100, n = 10, and x = 4.
a.
If the sample is drawn without replacement, the hypergeometric distribution should be used. The
hypergeometric distribution requires that sampling be done without replacement.
b.
If the sample is drawn with replacement, the binomial distribution should be used. The binomial
distribution requires that sampling be done with replacement.
With N = 10, n = 5, and r = 7, x can take on values 2, 3, 4, or 5.
a.
r N r 7 10 7
7! 3!
x n x 2 5 2
21(1)
P(x = 2) =
2!5! 3!0!
= .083
10!
252
N
10
n
5
5!5!
r N r 7 10 7
7! 3!
x n x 3 5 3
3!4!
2!1! 35(3) = .417
P(x = 3) =
10!
252
N
10
n
5
5!5!
r N r 7 10 7
7! 3!
x n x 4 5 4
35(3)
P(x = 4) =
4!3! 1!2!
= .417
10!
252
N
10
n
5
5!5!
r N r 7 10 7
7! 3!
x n x 5 5 5
5!2!
0!3! 21(1) = .083
P(x = 5) =
10!
252
N
10
n
5
5!5!
The probability distribution of x in tabular form is:
x
2
3
4
5
b.
μ=
σ2 =
p ( x)
.083
.417
.417
.083
nr 5(7)
= 3.5
N
10
r ( N r ) n( N n)
N ( N 1)
2
7(10 7)5(10 5)
10 (10 1)
2
525
= .583
900
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186
Chapter 4
c.
σ=
.5833 = .764
μ ± 2σ 3.5 ± 2(.764) 3.5 ± 1.528 (1.972, 5.028)
The graph of the distribution is:
0.4
p(x)
0.3
0.2
0.1
0.0
2
3
1.972
4.65
4.66
4
x
3.5
5
5.028
d.
P(1.972 < x < 5.028) = P(2 ≤ x ≤ 5) = 1.000
a.
For N = 209, r = 10, and n = 8, E ( x)
b.
8 209 8
8!
201!
4 10 4
4!(8 4)! 6!(201 6)!
P( x 4)
.0002
209!
209
10
10!(209 10)!
a.
E ( x) = μ = λ = 4
nr 10(8)
.383
N
209
σ λ 42
xμ
z
c.
Using Table III, Appendix B, with λ = 4,
σ
04
2.00
2
b.
P(x ≤ 10) = 1.000
4.67
a.
With λ = 4.5, P(x = 0) =
b.
P(x = 1) =
c.
μ = E(x) = λ = 4.5
4.50 e 4.5
0.0111
0!
4.51 e 4.5
0.0500
1!
σ λ 4.5 2.12
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Random Variables and Probability Distributions
4.68
187
Let x = number of male nannies in 10 trials. Then x is a binomial random variable with n = 10 and
p = 24 / 4,176 = .00575.
10
P( x 1) 1 P( x 0) 1 .005750.9942510 0 1 .9942510 1 .9440 .0560
0
4.69
Let x = number of “clean” cartridges selected in 5 trials. For this problem, N = 158, r = 122, and n = 5.
r N r 122 36
122! 36!
x n x 5 0
5!117! 0!36! .2693
P( x 5)
158!
N
158
n
5
5!153!
4.70
4.71
a.
Using Table III, Appendix B, with λ = 1.2, P(x = 0) = .301
b.
Using Table III, Appendix B, with λ = 1.2, P(x ≥ 2) = 1 – P(x ≤ 1) = 1 - .663 = .337
a.
Using Table III, Appendix B, with λ = 5,
P(x < 3) = P(x ≤ 2) = .125
4.72
b.
E(x) = λ = 5. The average number of calls blocked during the peak hour of video conferencing call
time is 5.
a.
Let x = number of defective items in a sample of size 4. For this problem, x is a hypergeometric
random variable with N = 10, n = 4, and r = 1. You will accept the lot if you observe no defectives.
r N r 1 10 1
1! 9!
x n x 0 4 0
0!1!
4!5! 1(126) = .6
P(x = 0) =
10!
210
N
10
n
4
4!6!
b.
If r = 2,
r N r 2 10 2
2! 8!
x n x 0 4 0
0!2!
4!4! 1(70) = .333
P(x = 0) =
10!
210
N
10
n
4
4!6!
4.73
Let x = number of spoiled bottles in the sample of 3. Since the sampling will be done without replacement,
x is a hypergeometric random variable with N = 12, n = 3, and r = 1.
r N r 1 12 1
1! 11!
x n x 1 3 1
1!0!
2!9! 55 .25
P(x = 1) =
12!
220
N
12
n
3
3!9!
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188
4.74
Chapter 4
a.
Using Table III and λ = 6.2, P(x = 2) = P(x ≤ 2) − P(x ≤ 1) = .054 − .015 = .039
P(x = 6) = P(x ≤ 6) − P(x ≤ 5) = .574 − .414 = .160
P(x = 10) = P(x ≤ 10) − P(x ≤ 9) = .949 − .902 = .047
b.
The plot of the distribution is:
c.
μ = λ = 6.2, σ = λ = 6.2 = 2.490
μ ± σ 6.2 ± 2.49 (3.71, 8.69)
μ ± 2σ 6.2 ± 2(2.49) 6.2 ± 4.98 (1.22, 11.18)
μ ± 3σ 6.2 ± 3(2.49) 6.2 ± 7.47 (−1.27, 13.67)
See the plot in part b.
d.
First, we need to find the mean number of customers per hour. If the mean number of customers per
10 minutes is 6.2, then the mean number of customers per hour is 6.2(6) = 37.2 = λ.
μ = λ = 37.2 and σ = λ = 37.2 = 6.099
μ ± 3σ 37.2 ± 3(6.099) 37.2 ± 18.297 (18,903, 55.498)
Using Chebyshev's Rule, we know at least 8/9 or 88.9% of the observations will fall within
3 standard deviations of the mean. The number 75 is way beyond the 3 standard deviation limit.
Thus, it would be very unlikely that more than 75 customers entered the store per hour on Saturdays.
4.75
a.
Using Table III, Appendix B, with λ = 10, P(x = 24) = P(x ≤ 24) – P(x ≤ 23) = 1.000 – 1.000 = .000
b.
Using Table III, Appendix B, with λ = 10, P(x = 23) = P(x ≤ 23) – P(x ≤ 22) = 1.000 – 1.000 = .000
c.
Yes, these probabilities are good approximations for the probability of “fire” and “theft”. The
researchers estimated these probabilities to be .0001, indicating that these would be extremely rare
events. Our probabilities of .000 are very close to .0001.
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Random Variables and Probability Distributions
4.76
189
μ = λ = 3, using Table III, Appendix B:
P(x = 0) = .050
The probability that no bulbs fail in one hour is .050. If we let y = number of one hour intervals out of 8
that have no bulbs fail, then y is a binomial random variable with n = 8 and p = .05. Then, the probability
that no bulbs fail in an 8 hour shift is
8
8!
.058.950
P(y = 8) = .058.95(88)
8!(8 8)!
8
=
8 7 6 5 4 3 2 1
.058.950 .058
8 7 6 5 4 3 2 11
We must assume that the 8 one-hour intervals are independent and identical, and that the probability that
no bulbs fail is the same for each one-hour interval.
4.77
Let x = number of females promoted in the 72 employees awarded promotion, where x is a
hypergeometric random variable. From the problem, N = 302, r = 73, and n = 72. We need to find if
observing 5 females who were promoted was fair.
E(x) = μ =
nr 72(73)
= 17.40
N
302
If 72 employees are promoted, we would expect that about 17 would be females.
V(x) = σ2 =
r ( N r ) n( N n)
N ( N 1)
2
73(302 73)72(302 72)
3022 (302 1)
= 10.084
σ = 10.084 = 3.176
Using Chebyshev’s Theorem, we know that at least 8/9 of all observations will fall within 3
standard deviations of the mean. The interval from 3 standard deviations below the mean to 3
standard deviations above the mean is:
μ ± 3σ 17.40 ± 3(3.176) 17.40 ± 9.528 (7.872, 26.928)
If there is no discrimination in promoting females, then we would expect between 8 and 26
females to be promoted within the group of 72 employees promoted. Since we observed only 5 females
promoted, we would infer that females were not promoted fairly.
4.78
If it takes exactly 5 minutes to wash a car and there are 5 cars in line, it will take 5(5) = 25
minutes to wash these 5 cars. Thus, for anyone to be in line at closing time, more than 1 car must
arrive in the final ½ hour. In addition, if on average 10 cars arrive per hour, then an average of 5
cars will arrive per ½ hour (30 minutes). If we let x = number of cars to arrive in ½ hour, then x
is a Poisson random variable with λ = 5.
P(x > 1) = 1 – P(x ≤ 1) = 1 − .04 = .96 (Using Table III, Appendix B)
Since this probability is so big, it is very likely that someone will be in line at closing time.
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Chapter 4
Using Table IV, Appendix B:
a.
P(z > 1.46) = .5 − P(0 < z ≤ 1.46)
= .5 − .4279 = .0721
b.
P(x <−1.56) = .5 − P(-1.56 ≤ z < 0)
= .5 − .4406 = .0594
c.
P(.67 ≤ z ≤ 2.41)
= P(0 < z ≤ 2.41) − P(0 < z < .67)
= .4920 − .2486 = .2434
d.
P(−1.96 ≤ z < −.33)
= P(−1.96 ≤ z < 0) − P(−.33 ≤ z < 0)
= .4750 − .1293 = .3457
e.
P(z ≥ 0) = .5
f.
P(−2.33 < z < 1.50)
= P(−2.33 < z < 0) + P(0 < z < 1.50)
= .4901 + .4332 = .9233
Table IV in the text gives the area between z = 0 and z = z0. In this exercise, the answers may thus be read
directly from the table by looking up the appropriate z.
a.
P(0 < z < 2.0) = .4772
b.
P(0 < z < 3.0) = .4987
c.
P(0 < z < 1.5) = .4332
d.
P(0 < z < .80) = .2881
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Random Variables and Probability Distributions
4.81
4.82
Using Table IV, Appendix B:
a.
P(−1 ≤ z ≤ 1)
= P(−1 ≤ z ≤ 0) + P(0 < z ≤ 1)
= .3413 + .3413 = .6826
b.
P(−1.96 ≤ z ≤ 1.96)
= P(−1.96 ≤ z < 0) + P(0 ≤ z ≤ 1.96)
= .4750 + .4750 = .9500
c.
P(−1.645 ≤ z ≤ 1.645)
= P(−1.645 ≤ z < 0) + P(0 ≤ z ≤ 1.645)
= .4500 + .4500 = .90
(using interpolation)
d.
P(−2 ≤ z ≤ 2)
= P(−2 ≤ z < 0) + P(0 ≤ z ≤ 2)
= .4772 + .4772 = .9544
a.
P(−1 ≤ z ≤ 1) = P(−1 ≤ z ≤ 0) + P(0 ≤ z ≤ 1)
= .3413 + .3413
= .6826
b.
P(−2 ≤ z ≤ 2) = P(−2 ≤ z ≤ 0) + P(0 ≤ z ≤ 2)
= .4772 + .4772
= .9544
c.
P(−2.16 < z ≤ 0.55) = P(−2.16 ≤ z ≤ 0) + P(0 ≤ z ≤ 0.55)
= .4846 + .2088
= .6934
d.
P(−.42 < z < 1.96)
= P(−.42 ≤ z ≤ 0) + P(0 ≤ z ≤ 1.96)
= .1628 + .4750
= .6378
e.
P(z ≥ −2.33) = P(−2.33 ≤ z ≤ 0) + P(z ≥ 0)
= .4901 + .5000
= .9901
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192
4.83
4.84
4.85
Chapter 4
f.
P(z < 2.33) = P(z ≤ 0) + P(0 ≤ z < 2.33)
= .5000 + .4901
= .9901
a.
P(z = 1) = 0, since a single point does not have an area.
b.
P(z ≤ 1) = P(z ≤ 0) + P(0 ≤ z ≤ 1)
= .5 + .3413
= .8413
(Table IV, Appendix B)
c.
P(z < 1) = P(z ≤ 1) = .8413 (Refer to part b.)
d.
P(z > 1) = 1 − P(z ≤ 1) = 1 − .8413 = .1587 (Refer to part b.)
Using Table IV, Appendix B:
a.
P(z ≥ z0) = .05
A1 = .5 − .05 = .4500
Looking up the area .4500 in Table IV gives
z0 = 1.645.
b.
P(z ≥ z0) = .025
A1 = .5 − .025 = .4750
Looking up the area .4750 in Table IV
gives z0 = 1.96.
c.
P(z ≤ z0) = .025
A1 = .5 − .025 = .4750
Looking up the area .4750 in Table IV gives
z = 1.96. Since z0 is to the left of 0, z0 = −1.96.
d.
P(z ≥ z0) = .10
A1 = .5 − .1 = .4
Looking up the area .4000 in Table IV
gives z0 = 1.28.
e.
P(z > z0) = .10
A1 = .5 − .1 = .4
z0 = 1.28 (same as in d)
Using Table IV of Appendix B:
a.
P(z ≤ z0) = .2090
A = .5000 − .2090 = .2910
Look up the area .2910 in the body of Table IV; z0 = −.81.
(z0 is negative since the graph shows z0 is on the left side of 0.)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
b.
P(z ≤ z0) = .7090
P(z ≤ z0) = P(z ≤ 0) + P(0 ≤ z ≤ z0)
= .5 + P(0 ≤ z ≤ z) = .7090
Therefore, P(0 ≤ z ≤ z0) = .7090 − .5 = .2090
Look up the area .2090 in the body of Table IV; z0 ≈ .55.
c.
P(−z0 ≤ z < z0) = .8472
P(−z0 ≤ z < z0) = 2P(0 ≤ z ≤ z0)
2P(0 ≤ z ≤ z0) = .8472
Therefore, P(0 ≤ z ≤ z0) = .4236.
Look up the area .4236 in the body of Table IV; z0 = 1.43.
d.
P(−z0 ≤ z < z0) = .1664
P(−z0 ≤ z ≤ z0) = 2P(0 ≤ z ≤ z0)
2P(0 ≤ z ≤ z0) = .1664
Therefore, P(0 ≤ z ≤ z0) = .0832.
Look up the area .0832 in the body of Table IV; z0 = .21.
e.
P(z0 ≤ z ≤ 0) = .4798
P(z0 ≤ z ≤ 0) = P(0 ≤ z ≤ −z0)
Look up the area .4798 in the body of Table IV;
z0 = −2.05.
f.
P(−1 < z < z0) = .5328
P(−1 < z < z0)
= P(−1 < z < 0) + P(0 < z < z0)
= .5328
P(0 < z < 1) + P(0 < z < z0) = .5328
Thus, P(0 < z < z0) = .5328 − .3413 = .1915
Look up the area .1915 in the body of Table IV; z0 = .50.
4.86
a.
z=1
b.
z = −1
c.
z=0
d.
z = −2.5
e.
z=3
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4.87
4.88
Chapter 4
Using Table IV, Appendix B:
a.
z=
b.
z=
c.
z=
d.
z=
e.
z=
f.
z=
xμ
σ
xμ
σ
xμ
σ
xμ
σ
xμ
σ
xμ
σ
20 30
= −2.50
4
30 30
=0
4
27.5 30
= −0.61
4
15 30
= −3.75
4
35 30
= 1.25
4
25 30
= −1.25
4
Using Table IV of Appendix B:
a.
To find the probability that x assumes a value more
than 2 standard deviations from μ:
P(x < μ − 2σ) + P(x > μ + 2σ)
= P(z < −2) + P(z > 2)
= 2P(z > 2)
= 2(.5000 − .4772)
= 2(.0228) = .0456
To find the probability that x assumes a value more than
3 standard deviations from μ:
P(x < μ − 3σ) + P(x > μ + 3σ)
= P(z < −3) + P(z > 3)
= 2P(z > 3)
= 2(.5000 − .4987)
= 2(.0013) = .0026
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Random Variables and Probability Distributions
b.
To find the probability that x assumes a value within 1
standard deviation of its mean:
P(μ − σ < x < μ + σ)
= P(−1 < z < 1)
= 2P(0 < z < 1)
= 2(.3413)
= .6826
To find the probability that x assumes a value within 2
standard deviations of μ:
P(μ − 2σ < x < μ + 2σ)
= P(−2 < z < 2)
= 2P(0 < z < 2)
= 2(.4772)
= .9544
c.
To find the value of x that represents the 80th percentile, we
must first find the value of z that corresponds to the 80th percentile.
P(z < z0) = .80. Thus, A1 + A2 = .80. Since A1 = .50,
A2 = .80 - .50 = .30. Using the body of Table IV, z0 = .84.
To find x, we substitute the values into the z-score formula:
z=
xμ
σ
.84 =
x 1000
x = .84(10) + 1000 = 1008.4
10
To find the value of x that represents the 10th percentile, we
must first find the value of z that corresponds to the 10th percentile.
P(z < z0) = .10. Thus, A1 = .50 - .10 = .40. Using the body
of Table IV, z0 = −1.28. To find x, we substitute the values
into the z-score formula:
z=
xμ
σ
−1.28 =
x 1000
x = −1.28(10) + 1000 = 987.2
10
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4.89
Chapter 4
a.
b.
c.
d.
e.
f.
12 11
10 11
z
P(10 ≤ x ≤ 12) = P
2
2
= P(−0.50 ≤ z ≤ 0.50)
= A1 + A2
= .1915 + .1915 = .3830
10 11
6 11
P(6 ≤ x ≤ 10) = P
z
2
2
= P(−2.50 ≤ z ≤ −0.50)
= P(−2.50 ≤ z ≤ 0) − P(−0.50 ≤ z ≤ 0)
= .4938 − .1915 = .3023
16 11
13 11
P(13 ≤ x ≤ 16) = P
z
2
2
= P(1.00 ≤ z ≤ 2.50)
= P(0 ≤ z ≤ 2.50) − P(0 ≤ x ≤ 1.00)
= .4938 − .3413 = .1525
P(7.8 ≤ x ≤ 12.6)
12.6 11
7.8 11
= P
z
2
2
= P(−1.60 ≤ z ≤ 0.80)
= A1 + A2
= .4452 + .2881 = .7333
13.24 11
P(x ≥ 13.24) = P z
2
= P(z ≥ 1.12)
= A2 = .5 − A1
= .5000 − .3686 = .1314
7.62 11
P(x ≥ 7.62) = P z
2
= P( z ≥ −1.69)
= A1 + A2
= .4545 + .5000 = .9545
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Random Variables and Probability Distributions
4.90
The random variable x has a normal distribution with μ = 50 and σ = 3.
a.
P(x ≤ x0) = .8413
So, A1 + A2 = .8413
Since A1 = .5, A2 = .8413 − .5 = .3413.
Look up the area .3413 in the body of Table IV,
Appendix B; z0 = 1.0.
To find x0, substitute all the values into the z-score formula:
z=
xμ
σ
x 50
1.0 = 0
3
x0 = 50 + 3(1.0) = 53
b.
P(x > x0) = .025
So, A = .5000 − .025 = .4750
Look up the area .4750 in the body of Table IV,
Appendix B; z0 = 1.96.
To find x0, substitute all the values into the z-score formula:
z=
xμ
σ
x 50
1.96 = 0
3
x0 = 50 + 3(1.96) = 55.88
c.
P(x > x0) = .95
So, A1 + A2 = .95. Since A2 = .5, A1 = .95 − .5 = .4500.
Look up the area .4500 in the body of Table IV,
Appendix B; (since it is exactly between two values,
average the z-scores). z0 ≈ −1.645.
To find x0, substitute into the z-score formula:
z=
xμ
σ
x 50
−1.645 = 0
3
x0 = 50 − 3(1.645) = 45.065
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Chapter 4
d.
P(41 ≤ x < x0) = .8630
z=
xμ
σ
41 50
= −3
3
A1 = P(41 ≤ x ≤ μ) = P(−3 ≤ z ≤ 0)
= P(0 ≤ z ≤ 3)
= .4987
A1 + A2 = .8630, since A1 = .4987, A2 = .8630 - .4987 = .3643. Look up .3643 in the body of Table
IV, Appendix B; z0 = 1.1.
To find x0, substitute into the z-score formula:
z=
xμ
σ
x 50
1.1 = 0
3
x0 = 50 + 3(1.1) = 53.3
e.
P(x < x0) = .10
So A = .5000 − .10 = .4000
Look up area .4000 in the body of Table IV, Appendix B; z0 = 1.28. Since z0 is to the left of 0,
z0 = −1.28.
To find x0, substitute all the values into the z-score formula:
z=
xμ
σ
x0 50
3
x0 = 50 − 1.28(3) = 46.16
−1.28 =
f.
P(x > x0) = .01
So A = .5000 − .01 = .4900
Look up area .4900 in the body of Table IV, Appendix B; z0 = 2.33.
To find x0, substitute all the values into the z-score formula:
z=
xμ
σ
x0 50
3
x0 = 50 + 2.33(3) = 56.99
2.33 =
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.91
Let x = age of a powerful woman. The random variable x has a normal distribution with μ = 50 and σ =
5.3. Using Table IV, Appendix B,
a.
b.
c.
d.
4.92
a.
60 50
55 50
P(55 x 60) P
z
P(.83 z 1.67)
6
6
P(0 z 1.67) P(0 z .83) .4525 .2967 .1558
52 50
48 50
P(48 x 52) P
z
P(.33 z .33)
6
6
P(.33 z 0) P (0 z .33) .1293 .1293 .2586
35 50
P( x 35) P z
P( z 2.5)
6
.5 P (2.5 z 0) .5 .4933 .0062
40 50
P( x 40) P z
P( z 1.67)
6
.5 P (1.67 z 0) .5 .4525 .9525
Using Table IV, Appendix B,
0 5.26
P( x 0) P z
P( z 0.526)
10
.5 P(0.53 z 0) .5 .2019 .7019
b.
15 5.26
5 5.26
P(5 x 15) P
z
P (0.026 z 0.974)
10
10
P(.03 z 0) P(0 z .97) .0120 .3340 .3460
c.
1 5.26
P( x 1) P z
P ( z 0.426)
10
.5 P(0.43 z 0) .5 .1664 .3336
d.
25 5.26
P( x 25) P z
P ( z 3.026)
10
.5 P(3.03 z 0) .5 .4988 .0012
Since the probability of seeing a win percentage of -25% or anything more unusual is so small (p =
.0012), we would conclude that the average casino win percentage is not 5.26%.
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200
4.93
Chapter 4
a.
Let x = buy-side analyst’s forecast error. Then x has an approximate normal distribution with
µ = .85 and σ = 1.93. Using Table IV, Appendix B,
2.00 .85
P( x 2.00) P z
P ( z .60) .5 .2257 .2743
1.93
b.
Let y = sell-side analyst’s forecast error. Then y has an approximate normal distribution with
µ = -.05 and σ = 85. Using Table IV, Appendix B,
2.00 (.05)
P( y 2.00) P z
P( z 2.41) .5 .4920 .0080
.85
4.94
Let x = driver’s head injury rating. The random variable x has a normal distribution with
μ = 605 and σ = 185. Using Table IV, Appendix B,
a.
b.
c.
d.
4.95
700 605
500 605
P(500 x 700) P
z
P (0.57 z 0.51)
185
185
P(0.57 z 0) P(0 z 0.51) .2157 .1950 .4107
500 605
400 605
P(400 x 500) P
z
P(1.11 z 0.57)
185
185
P(1.11 z 0) P(0.57 z 0) .3665 .2157 .1508
850 605
P( x 850) P z
P( z 1.32) .5 P (0 z 1.32)
185
.5 .4066 .9066
1, 000 605
P( x 1, 000) P z
P( z 2.14) .5 P (0 z 2.14)
185
.5 .4838 .0162
Let x = transmission delay. The random variable x has a normal distribution with μ = 48.5 and σ = 8.5.
Using Table IV, Appendix B,
a.
b.
57 48.5
P( x 57) P z
P( z 1.00)
8.5
.5 P (0 z 1) .5 .3413 .8413
60 48.5
40 48.5
P(40 x 60) P
z
P (1 z 1.35)
8.5
8.5
P(1 z 0) P(0 z 1.35) .3413 .4115 .7528
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Random Variables and Probability Distributions
4.96
201
Let x = number of defects per million. Then x has an approximate normal distribution with µ = 3. Using
Table IV, Appendix B,
3 1.5σ 3
3 1.5σ 3
P(3 1.5σ x 3 1.5σ ) P
z
P(1.5 z 1.5) .4332 .4332 .8664
σ
σ
It is fairly likely that the goal will b met. Since the probability is .8664, the goal would be met
approximately 86.64% of the time.
4.97
a.
Let x = rating. Then x has a normal distribution with µ = 50 and σ = 15. Using Table IV,
Appendix B,
P(x > xo) = .10. Find xo.
x 50
P( x xo ) P z o
P ( z zo ) .10
15
A1 = .5 − .10 = .4000
Looking up area .4000 in Table IV, zo = 1.28
zo
b.
xo 50
x 50
1.28 o
xo 50 1.28(15) 69.2
15
15
P(x > xo) = .10 + .20 + .40 = .70. Find xo.
x 50
P( x xo ) P z o
P ( z zo ) .70
15
A1 = .70 − .5 = .2000
Looking up area .2000 in Table IV, zo = −.52
zo
4.98
a.
xo 50
x 50
.52 o
xo 50 .52(15) 42.2
15
15
Let x = crop yield. The random variable x has a normal distribution with μ = 1,500
and σ = 250.
1, 600 -1,500
P(x < 1,600) = P z
= P(z < .4) = .5 + .1554 = .6554
250
(Using Table IV)
b.
Let x1 = crop yield in first year and x2 = crop yield in second year. If x1 and x2 are independent, then
the probability that the farm will lose money for two straight years is:
1, 600 1,500
1, 600 1,500
P(x1 < 1,600) P(x2 < 1,600) = P z1
P z2
250
250
= P(z1 < .4) P(z2 < .4) = (.5 + .1554)(.5 + .1554) = .6554(.6554) = .4295
(Using Table IV)
c.
[1,500 2σ ] 1,500
[1,500 2σ ] 1,500
z
P(1,500 − 2σ ≤ x ≤ 1,500 + 2σ) = P
σ
σ
(Using Table IV)
= P(−2 ≤ z ≤ 2) = 2P(0 ≤ z ≤ 2) = 2(.4772) = .9544
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202
4.99
Chapter 4
a.
Using Table IV, Appendix B, and μ = 75 and σ = 7.5,
80 75
P(x > 80) = P z
= P(z > .67) = .5 - .2486 = .2514
7.5
Thus, 25.14% of the scores exceeded 80.
b.
P(x ≤ x0) = .98. Find x0.
x 75
P(x ≤ x0) = P z 0
= P(z ≤ z0) = .98
7.5
A1 = .98 - .5 = .4800
Looking up area .4800 in Table IV, z0 = 2.05.
z0 =
4.100
x 0 75
x 75
2.05 = 0
x0 = 90.375
7.5
7.5
Let x = wage rate. The random variable x is normally distributed with μ = 17 and σ = 1.25. Using Table
IV, Appendix B,
a.
18.30 17
P( x 18.30) P z
P( z 1.04)
1.25
.5 P(0 z 1.04) .5 .3508 .1492
b.
18.30 17
P( x 18.30) P z
P( z 1.04)
1.25
.5 P(0 z 1.04) .5 .3508 .1492
c.
P(x ≤ η) = P(x ≥ η) = .5
Thus, μ = η = 17.
(Recall from section 2.4 that in a symmetric distribution, the mean equals the median.)
4.101
a.
The contract will be profitable if total cost, x, is less than $1,000,000.
1, 000, 000 850, 000
P(x < 1,000,000) = P z
= P(z < .88) = .5 + .3106 = .8106
170, 000
b.
The contract will result in a loss if total cost, x, exceeds 1,000,000.
P(x > 1,000,000) = 1 − P(x < 1,000,000) = 1 − .8106 = .1894
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Random Variables and Probability Distributions
c.
203
P(x < R) = .99. Find R.
R 850, 000
= P(z < z0) = .99
P(x < R) = P z
170, 000
A1 = .99 − .5 = .4900
Looking up the area .4900 in Table IV, z0 = 2.33
R 850, 000
R 850, 000
2.33 =
170, 000
170, 000
R = 2.33(170,000) + 850,000 = $1,246,100
z0 =
4.102
a.
Let x = quantity injected per container. The random variable x has a normal distribution with μ = 10
and σ = .2.
10 10
P(x < 10) = P z
= P(z < 0.0) = .5
.2
10 10
P(x ≥ 10) = P z
= P(z ≥ 0.0) = .5
.2
4.103
b.
Since the container needed to be reprocessed, it cost $10. Upon refilling, it contained 10.60 units
with a cost of 10.60($20) = $212. Thus, the total cost for filling this container is $10 + $212 = $222.
Since the container sells for $230, the profit is $230 − $222 = $8.
c.
Let x = quantity injected per container. The random variable x has a normal distribution with μ =
10.10 and σ = .2. The expected value of x is E(x) = μ = 10.10. The cost of a container with 10.10
units is 10.10($20) = $202. Thus, the expected profit would be the selling price minus the cost or
$230 − $202 = $28.
Let x = load. Then x has a normal distribution with µ = 20. We are given P(10 < x < 30) = .95. We want
to find σ.
P(10 < x < 30) = .95 P(z1 < z < z2) = .95 P(z1 < z < 0) = P(0 < z < z2) = .95/2 = .4750
Looking up area .4750 in Table IV, Appendix B, z2 = 1.96 and z1 = −1.96.
z2
4.104
a.
x 30
σ
30 20
σ
1.96 σ
10
5.1
1.96
If z is a standard normal random variable,
QL = zL is the value of the standard normal distribution which has 25% of the data to the left and 75%
to the right.
Find zL such that P(z < zL) = .25
A1 = .50 − .25 = .25.
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204
Chapter 4
Look up the area A1 = .25 in the body of Table IV of Appendix B; zL = −.67 (taking the closest
value). If interpolation is used, −.675 would be obtained.
QU = zU is the value of the standard normal distribution which has 75% of the data to the left and
25% to the right.
Find zU such that P(z < zU) = .75
A1 + A2 = P(z ≤ 0) + P(0 ≤ z ≤ zU)
= .5 + P(0 ≤ z ≤ zU)
= .75
Therefore, P(0 ≤ z ≤ zU) = .25.
Look up the area .25 in the body of Table IV of Appendix B; zU = .67 (taking the closest value).
b.
Recall that the inner fences of a box plot are located 1.5(QU − QL) outside the hinges (QL and QU).
To find the lower inner fence,
QL − 1.5(QU − QL) = −.67 − 1.5(.67 − (−.67))
= −.67 − 1.5(1.34)
= −2.68 (−2.70 if zL = −.675 and zU = +.675)
The upper inner fence is:
QU + 1.5(QU − QL) = .67 + 1.5(.67 − (−.67))
= .67 + 1.5(1.34)
= 2.68 (+2.70 if zL = −.675 and zU = +.675)
c.
Recall that the outer fences of a box plot are located 3(QU − QL) outside the hinges
(QL and QU).
To find the lower outer fence,
QL − 3(QU − QL) = −.67 − 3(.67 − (−.67))
= −.67 − 3(1.34)
= −4.69 (−4.725 if zL = −.675 and zU = +.675)
The upper outer fence is:
QU + 3(QU − QL) = .67 + 3(.67 − (−.67))
= .67 + 3(1.34)
= 4.69 (4.725 if zL = −.675 and zU = +.675)
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Random Variables and Probability Distributions
d.
205
P(z < −2.68) + P(z > 2.68)
= 2P(z > 2.68)
= 2(.5000 − .4963)
(Table IV, Appendix B)
= 2(.0037) = .0074
(or 2(.5000 − .4965) = .0070 if −2.70 and 2.70 are used)
P(z < −4.69) + P(z > 4.69)
= 2P(z > 4.69)
≈ 2(.5000 − .5000) ≈ 0
4.105
4.106
e.
In a normal probability distribution, the probability of an observation being beyond the inner fences
is only .0074 and the probability of an observation being beyond the outer fences is approximately
zero. Since the probability is so small, there should not be any observations beyond the inner and
outer fences. Therefore, they are probably outliers.
a.
The proportion of measurements that one would expect to fall in the interval μ ± σ is about .68.
b.
The proportion of measurements that one would expect to fall in the interval μ ± 2σ is about .95.
c.
The proportion of measurements that one would expect to fall in the interval μ ± 3σ is about 1.00.
a.
IQR = QU − QL = 195 − 72 = 123
b.
IQR/s = 123/95 = 1.295
c.
Yes. Since IQR is approximately 1.3, this implies that the data are approximately normal.
4.107
If the data are normally distributed, then the normal probability plot should be an approximate straight line.
Of the three plots, only plot c implies that the data are normally distributed. The data points in plot c form
an approximately straight line. In both plots a and b, the plots of the data points do not form a straight line.
4.108
a.
Using MINITAB, the stem-and-leaf display is:
Stem-and-leaf of X
N = 28
Leaf Unit = 0.10
5
6
8
11
14
14
10
7
2
2
3
4
5
6
7
8
11266
1
35
035
039
3457
346
24469
47
Since the data do not form a mound-shape, it indicates that the data may not be normally distributed.
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Chapter 4
b.
Using MINITAB, the descriptive statistics are:
Variable
N
Mean
Median
TrMean
StDev
SE Mean
X
28
5.511
6.100
5.519
2.765
0.5230
Variable
X
Minimum
Maximum
Q1
Q3
1.100
9.700
3.350
8.050
The standard deviation is 2.765.
c.
Using the printout from MINITAB in part b, QL = 3.35, and QU = 8.05. The IQR
= QU − QL = 8.05 − 3.35 = 4.7. If the data are normally distributed, then IQR/s ≈ 1.3.
For this data, IQR/s = 4.7/2.765 = 1.70. This is a fair amount larger than 1.3, which indicates that the
data may not be normally distributed.
d.
Using MINITAB, the normal probability plot is:
The data at the extremes are not particularly on a straight line. This indicates that the data are not
normally distributed.
4.109
a.
IQR = QU – QL = 54 – 47 = 7
This agrees with IQR from the printout.
b.
From the printout, s = 6.444
c.
If the data are approximately normal, then
IQR
1.3 . For this problem,
s
IQR
7
1.086 1.3 . Thus, the distribution is approximately normal.
s
6.444
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Random Variables and Probability Distributions
d.
207
From Exercise 2.48 d, the histogram is:
Histogram of AGE
14
12
Frequency
10
8
6
4
2
0
32
40
48
56
64
AGE
From the histogram, the data appear to be approximately mound-shaped. Thus, the data are
approximately normal.
4.110
Based on the normal probability plot, it appears that the data are not approximately normal. If
the data are normal, then the probability plot should reflect a straight line. In this graph, the plot of the data
is not a straight line.
4.111
The information given in the problem states that x = 4.71, s = 6.09, QL = 1, and QU = 6. To be normal, the
data have to be symmetric. If the data are symmetric, then the mean would equal the median and would be
half way between the lower and upper quartile. Half way between the upper and lower quartiles is 3.5.
The sample mean is 4.71, which is much larger than 3.5. This implies that the data may not be normal. In
addition, the interquartile range divided by the standard deviation will be approximately 1.3 if the data are
normal. For this data,
IQR QU QL 6 1
.82
s
6.09
s
The value of .82 is much smaller than the necessary 1.3 to be normal. Again, this is an indication that the
data are not normal. Finally, the standard deviation is larger than the mean. Since one cannot have values
of the variable in this case less than 0, a standard deviation larger than the mean indicates that the data are
skewed to the right. This implies that the data are not normal.
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208
We will look at the 4 methods or determining if the data are normal. First, we will look at a
histogram of the data. Using MINITAB, the histogram of the failure times of the 50 used panels is:
Histogram of Fail
Normal
12
Mean
StDev
N
10
1.935
0.9287
50
8
Frequency
4.112
Chapter 4
6
4
2
0
0
1
2
Fail
3
4
From the histogram, the data appear to have a somewhat normal distribution.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB,
the summary statistics are:
Descriptive Statistics: Fail
Variable
Fail
N
50
Mean
1.935
StDev
0.929
Q1
1.218
Median
1.835
Q3
2.645
x s 1.935 .929 (1.006, 2.864) 33 of the 50 values fall in this interval. The proportion is
33/50 = .66. This is fairly close to the .68 we would expect if the data were normal.
x 2s 1.935 2(.929) 1.935 1.858 (0.077, 3.793) 49 of the 50 values fall in this interval. The
proportion is 49/50 = .98. This is a fair amount above the .95 we would expect if the data were normal.
x 3s 1.935 3(.929) 1.935 2.787 (0.852, 4.722) 50 of the 50 values fall in this interval.
The proportion is 50/50 =1.00. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the data may be normal.
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 2.645 – 1.218 = 1.427.
IQR 1.427
1.54 . This is somewhat larger than the 1.3 we would expect if the data were normal. This
s
.929
method indicates the data may be normal.
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Random Variables and Probability Distributions
209
Finally, using MINTAB, the normal probability plot is:
Probability Plot of Fail
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
1.935
0.9287
50
0.305
0.557
Percent
80
70
60
50
40
30
20
10
5
1
-1
0
1
2
Fail
3
4
5
Since the data form a fairly straight line, the data may be normal.
From the 4 different methods, all indications are that the failure times are approximately normal.
We will look at the 4 methods for determining if the data are normal. First, we will look at a
histogram of the data. Using MINITAB, the histogram of the driver’s head injury rating is:
20
Frequency
4.113
10
0
200
300 400
500 600
700
800 900 1000 1100 1200
DrivHead
From the histogram, the data appear to be somewhat skewed to the right, but is fairly mound-shaped.
This indicates that the data are normal.
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Chapter 4
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB,
the summary statistics are:
Descriptive Statistics: DrivHead
Variable
DrivHead
N
98
Mean
603.7
Median
605.0
TrMean
600.3
Variable
DrivHead
Minimum
216.0
Maximum
1240.0
Q1
475.0
Q3
724.3
StDev
185.4
SE Mean
18.7
x s 603.7 185.4 (418.3, 789.1) 68 of the 98 values fall in this interval. The proportion is .69.
This is very close to the .68 we would expect if the data were normal.
x 2s 603.7 2(185.4) 603.7 370.8 (232.9, 974.5) 96 of the 98 values fall in this interval.
The proportion is .98. This is a fair amount larger than the .95 we would expect if the data were normal.
x 3s 603.7 3(185.4) 603.7 556.2 (47.5, 1,159.9) 97 of the 98 values fall in this interval.
The proportion is .99. This is fairly close to the 1.00 we would expect if the data were normal.
From this method, it appears that the data may be normal.
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 724.3 – 475 = 249.3.
IQR 249.3
1.3 This is equal to the 1.3 we would expect if the data were normal. This method
185.4
s
indicates the data may be normal.
Finally, using MINITAB, the normal probability plot is:
Since the data form a fairly straight line, the data may be normal.
From the 4 different methods, all indications are that the driver’s head injury rating data are normal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of
the data. Using MINITAB, the histogram of the sanitation scores is:
Histogram of Score
60
50
40
Frequency
4.114
211
30
20
10
0
60.0
67.5
75.0
Score
82.5
90.0
97.5
From the histogram, the data appear to be skewed to the left. This indicates that the data are not normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB,
the summary statistics are:
Descriptive Statistics: DDT
Variable
Score
N
182
Mean
95.044
StDev
5.391
Minimum
56.000
Q1
94.000
Median
96.500
Q3
98.000
Maximum
100.000
x s 95.044 5.391 (89.653, 100.435) 164 of the 182 values fall in this interval. The proportion is
.90. This is much larger than the .68 we would expect if the data were normal.
x 2s 95.044 2(5.391) 95.044 10.782 (84.262, 105.826) 174 of the 182 values fall in this
interval. The proportion is .96. This is slightly larger than the .95 we would expect if the data were
normal.
x 3s 95.044 3(5.391) 95.044 16.173 (78.871, 111.271) 178 of the 182 values fall in this
interval. The proportion is .978. This is somewhat smaller than the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 98 – 94 = 4.
IQR
4
.742 This is much smaller than the 1.3 we would expect if the data were normal. This
s
5.391
method indicates the data are not normal.
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212
Chapter 4
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Score
Normal - 95% CI
99.9
Mean
StDev
N
AD
P-Value
99
Percent
95
90
95.04
5.391
182
10.307
<0.005
80
70
60
50
40
30
20
10
5
1
0.1
50
60
70
80
90
100
110
120
Score
Since the data do not form a straight line, the data are not normal.
From the 4 different methods, all indications are that the sanitation scores data are not normal.
We will look at the 4 methods or determining if the 3 variables are normal.
Distance:
First, we will look at A histogram of the data. Using MINITAB, the histogram of the distance data is:
Histogram of DISTANCE
Normal
18
Mean
StDev
N
16
298.9
7.525
40
14
12
Frequency
4.115
10
8
6
4
2
0
285
290
295
300
305
DISTANCE
310
315
320
From the histogram, the distance data do not appear to have a normal distribution.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB,
the summary statistics are:
Descriptive Statistics: DISTANCE, ACCURACY, INDEX
Variable
DISTANCE
N
40
Mean
298.95
StDev
7.53
Minimum
283.20
Q1
294.60
Median
299.05
Q3
302.00
Maximum
318.90
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Random Variables and Probability Distributions
213
x s 298.95 7.53 (291.42, 306.48) 28 of the 40 values fall in this interval. The proportion is
28/40 = .70. This is fairly close to the .68 we would expect if the data were normal.
x 2s 298.95 2(7.53) 298.95 15.06 (283.89, 314.01) 37 of the 40 values fall in this interval.
The proportion is 37/40 = .925. This is a fair amount below the .95 we would expect if the data were
normal.
x 3s 298.95 3(7.53) 298.95 22.59 (276.36, 321.54) 40 of the 40 values fall in this interval.
The proportion is 40/40 =1.00. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the distance data may not be normal.
Next, we look at the ratio of the IQR to s.
IQR = QU – QL = 302 – 294.6 = 7.4.
IQR
7.4
.983 . This is much smaller than the 1.3 we would expect if the data were normal. This
s
7.53
method indicates the distance data may not be normal.
Finally, using MINTAB, the normal probability plot is:
Probability Plot of DISTANCE
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
298.9
7.525
40
0.521
0.174
Percent
80
70
60
50
40
30
20
10
5
1
280
290
300
DISTANCE
310
320
Since the data do not form a fairly straight line, the distance data may not be normal.
From the 4 different methods, all indications are that the distance data are not normal.
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214
Chapter 4
Accuracy:
First, we will look at a histogram of the data. Using MINITAB, the histogram of the accuracy data is:
Histogram of ACCURACY
Normal
14
Mean
StDev
N
12
61.97
5.226
40
Frequency
10
8
6
4
2
0
48
54
60
ACCURACY
66
72
From the histogram, the accuracy data do not appear to have a normal distribution.
Descriptive Statistics: DISTANCE, ACCURACY, INDEX
Variable
ACCURACY
N
40
Mean
61.970
StDev
5.226
Minimum
45.400
Q1
59.400
Median
61.950
Q3
64.075
Maximum
73.000
x s 61.97 5.226 (56.744, 67.196) 30 of the 40 values fall in this interval. The proportion is
30/40 = .75. This is much greater than the .68 we would expect if the data were normal.
x 2s 61.97 2(5.226) 61.97 10.452 (51.518, 72.422) 37 of the 40 values fall in this interval.
The proportion is 37/40 = .925. This is a fair amount below the .95 we would expect if the data were
normal.
x 3s 61.97 3(5.226) 61.97 15.678 (46.292, 77.648) 39 of the 40 values fall in this interval.
The proportion is 39/40 =.975. This is a fair amount lower than the 1.00 we would expect if the data were
normal.
From this method, it appears that the accuracy data may not be normal.
Next, we look at the ratio of the IQR to s.
IQR = QU – QL = 64.075 – 59.4 = 4.675.
IQR 4.675
.895 . This is much smaller than the 1.3 we would expect if the data were normal. This
s
5.226
method indicates the accuracy data may not be normal.
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Random Variables and Probability Distributions
215
Finally, using MINTAB, the normal probability plot is:
Probability Plot of ACCURACY
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
61.97
5.226
40
0.601
0.111
Percent
80
70
60
50
40
30
20
10
5
1
45
50
55
60
65
ACCURACY
70
75
80
Since the data do not form a fairly straight line, the accuracy data may not be normal.
From the 4 different methods, all indications are that the accuracy data are not normal.
Index:
First, we will look at a histogram of the data. Using MINITAB, the histogram of the index data is:
Histogram of INDEX
Normal
Mean
StDev
N
10
1.927
0.6602
40
Frequency
8
6
4
2
0
0.5
1.0
1.5
2.0
INDEX
2.5
3.0
3.5
From the histogram, the index data do not appear to have a normal distribution.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB,
the summary statistics are:
Descriptive Statistics: DISTANCE, ACCURACY, INDEX
Variable
INDEX
N
40
Mean
1.927
StDev
0.660
Minimum
1.170
Q1
1.400
Median
1.755
Q3
2.218
Maximum
3.580
x s 1.927 .660 (1.267, 2.587) 30 of the 40 values fall in this interval. The proportion is 30/40 =
.75. This is much greater than the .68 we would expect if the data were normal.
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216
Chapter 4
x 2s 1.927 2(.660) 1.927 1.320 (.607, 3.247) 37 of the 40 values fall in this interval. The
proportion is 37/40 = .925. This is a fair amount below the .95 we would expect if the data were normal.
x 3s 1.927 3(.660) 1.927 1.980 (.053, 3.907) 40 of the 40 values fall in this interval. The
proportion is 40/40 = 1.000. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the index data may not be normal.
Next, we look at the ratio of the IQR to s.
IQR = QU – QL = 2.218 – 1.4 = .818.
IQR .818
1.23 . This is fairly close to the 1.3 we would expect if the data were normal. This method
s
.66
indicates the index data may normal.
Finally, using MINTAB, the normal probability plot is:
Probability Plot of INDEX
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
1.927
0.6602
40
1.758
<0.005
Percent
80
70
60
50
40
30
20
10
5
1
0
1
2
INDEX
3
4
Since the data do not form a fairly straight line, the index data may not be normal.
From 3 of the 4 different methods, the indications are that the index data are not normal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of
the data. Using MINITAB, the histogram of the tensile strength values is:
Histogram of Strength
3.0
2.5
Fr equency
4.116
217
2.0
1.5
1.0
0.5
0.0
330
335
340
345
Str ength
350
355
From the histogram, the data appear to be somewhat skewed to the left. This might indicate that the data
are not normal.
Next, we look at the intervals x s, x 2s, x 3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB,
the summary statistics are:
Descriptive Statistics: Strength
Variable
Strength
N
11
N*
0
Variable
Strength
Maximum
356.30
Mean
342.13
SE Mean
2.38
StDev
7.91
Minimum
328.20
Q1
334.70
Median
343.60
Q3
347.80
x s 342.13 7.91 (334.22, 350.04) 8 of the 11 values fall in this interval. The proportion is .73.
This is somewhat larger than the .68 we would expect if the data were normal.
x 2s 342.16 2(7.91) 342.16 9.65 (326.34, 357.98) All 11 of the 11 values fall in this
interval. The proportion is 1.00. This is somewhat larger than the .95 we would expect if the data were
normal.
x 3s 342.16 3(7.91) 342.16 23.73 (318.43, 365.89) Again, all 11 of the 11 values fall in this
interval. The proportion is 1.00. This is equal to the 1.00 we would expect if the data were normal.
From this method, it appears that the data are quite normal.
Next, we look at the ratio of the IQR to s. IQR = QU – QL = 347.80 – 334.70 = 13.1.
IQR 13.1
1.656 This is much larger than the 1.3 we would expect if the data were normal. This
s
7.91
method indicates the data are not normal.
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218
Chapter 4
Finally, using MINITAB, the normal probability plot is:
Probability Plot of Strength
Normal - 95% CI
99
Mean
StDev
N
AD
P-Value
95
90
342. 1
7.907
11
0.154
0.937
80
Percent
70
60
50
40
30
20
10
5
1
310
320
330
340
Strength
350
360
370
Since the data do form a fairly straight line, the data could be normal.
From the 4 different methods, three of the four indicate that the data probably are not from a normal
distribution.
4.117
From Exercise 2.51, it states that the mean number of semester hours for those taking the CPA exam is
141.31 and the median is 140. It also states that most colleges only require 128 semester hours for an
undergraduate degree. Thus, the minimum value for the total semester hours is around 128. The
z-score associated with 128 is:
z
xμ
σ
128 141.31
.75
17.77
If the data are normal, we know that about .34 of the observations are between the mean and 1 standard
deviation below the mean. Thus, .16 of the observations are more than 1 standard deviation below the
mean. With this distribution, that is impossible. Thus, the data are not normal. The mean is greater than
the median, so we know that the data are skewed to the right.
4.118
a.
In order to approximate the binomial distribution with the normal distribution, the interval μ ± 3σ
np ± 3 npq should lie in the range 0 to n.
When n = 25 and p = .4,
np ± 3 npq 25(.4) ± 3 25(.4)(1 .4)
10 ± 3 6 10 ± 7.3485 (2.6515, 17.3485)
Since the interval calculated does lie in the range 0 to 25, we can use the normal approximation.
b.
μ = np = 25(.4) = 10
σ2 = npq = 25(.4)(.6) = 6
c.
P(x ≥ 9) = 1 − P(x ≤ 8) = 1 − .274 = .726
(Table II, Appendix B)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
d.
4.119
a.
(9 .5) 10
P(x ≥ 9) ≈ P z
6
= P(z ≥−.61)
= .5000 + .2291 = .7291
(Using Table IV in Appendix B.)
μ = np = 100(.01) = 1.0, σ =
npq 100(.01)(.99) = .995
μ ± 3σ 1 ± 3(.995) 1 ± 2.985 (−1.985, 3.985)
Since this interval does not fall in the interval (0, n = 100), the normal approximation is not
appropriate.
b.
μ = np = 20(.6) = 12, σ =
npq 20(.6)(.4) = 2.191
μ ± 3σ 12 ± 3(2.191) 12 ± 6.573 (5.427, 18.573)
Since this interval falls in the interval (0, n = 20), the normal approximation is appropriate.
c.
μ = np = 10(.4) = 4, σ =
npq 10(.4)(.6) = 1.549
μ ± 3σ 4 ± 3(1.549) 4 ± 4.647 (−.647, 8.647)
Since this interval does not fall within the interval (0, n = 10), the normal approximation is not
appropriate.
d.
μ = np = 1000(.05) = 50, σ =
npq 1000(.05)(.95) = 6.892
μ ± 3σ 50 ± 3(6.892) 50 ± 20.676 (29.324, 70.676)
Since this interval falls within the interval (0, n = 1000), the normal approximation is appropriate.
e.
μ = np = 100(.8) = 80, σ =
npq 100(.8)(.2) = 4
μ ± 3σ 80 ± 3(4) 80 ± 12 (68, 92)
Since this interval falls within the interval (0, n = 100), the normal approximation is appropriate.
f.
μ = np = 35(.7) = 24.5, σ =
npq 35(.7)(.3) = 2.711
μ ± 3σ 24.5 ± 3(2.711) 24.5 ± 8.133 (16.367, 32.633)
Since this interval falls within the interval (0, n = 35), the normal approximation is appropriate.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
219
220
4.120
Chapter 4
μ = np = 1000(.5) = 500, σ = npq 1000(.5)(.5) = 15.811
a.
Using the normal approximation,
(500 .5) 500
P(x > 500) ≈ P z
= P(z > .03) = .5 − .0120 = .4880
15.811
(from Table IV, Appendix B)
b.
(500 .5) 500
(490 .5) 500
z
P(490 ≤ x < 500) ≈ P
15.811
15.811
= P(−.66 ≤ z < −.03) = .2454 − .0120 = .2334
(from Table IV, Appendix B)
c.
4.121
(550 .5) 500
P(x > 550) ≈ P z
= P(z > 3.19) ≈ .5 − .5 = 0
15.811
(from Table IV, Appendix B)
x is a binomial random variable with n = 100 and p = .4.
μ ± 3σ np ± 3 npq 100(.4) ± 3 100(.4)(1 .4)
40 ± 3(4.8990) (25.303, 54.697)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probabilities.
a.
b.
c.
(35 .5) 40
P(x ≤ 35) ≈ P z
4.899
= P(z ≤ −.92)
= .5000 −.3212 = .1788
(Using Table IV in Appendix B.)
P(40 ≤ x ≤ 50)
(50 .5) 40
(40 .5) 40
z
≈ P
4.899
4.899
= P(−.10 ≤ z ≤ 2.14)
= P(−.10 ≤ z ≤ 0) + P(0 ≤ z ≤ 2.14)
= .0398 + .4838 = .5236
(Using Table IV in Appendix B.)
(38 .5) 40
P(x ≥ 38) ≈ P z
4.899
= P(z ≥ −.51)
= .5000 + .1950 = .6950
(Using Table IV in Appendix B.)
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Random Variables and Probability Distributions
4.122
a.
x is a binomial random variable with n = 100 and p = .2.
b.
µ = E(x) = np = 100(.20) = 20
σ npq 100(.2)(.8) 16 4
c.
z
d.
4.123
xμ
σ
27.5 20
1.875
4
25 .5 20
P( x 25) P z
P( z 1.38) .5 .4162 .9162
4
(Using Table IV, Appendix B)
Let x = number of patients who undergo laser surgery who have serious post-laser vision problems in
100,000 trials. Then x is a binomial random variable with n = 100,000 and p = .01.
E(x) = μ = np = 100,000(.01) = 1,000.
σ σ 2 npq 100, 000(.01)(.99) 990 31.464
To see if the normal approximation is appropriate, we use:
μ 3σ 1, 000 3(31.464) 1, 000 94.392 (905.608, 1, 094.392)
Since the interval lies in the range of 0 to 100,000, the normal approximation is appropriate.
949.5 1000
P( x 950) P z
P ( z 1.61) .5 .4463 .0537
31.464
(Using Table IV, Appendix B)
4.124
a.
E(x) = μ = np = 350(.27) = 94.5.
b.
σ σ 2 npq 350(.27)(.73) 68.985 8.306
c.
z
d.
To see if the normal approximation is appropriate, we use:
xμ
σ
99.5 94.5
0.60
8.306
μ 3σ 94.5 3(8.306) 94.5 24.918 (69.582, 119.418)
Since the interval lies in the range of 0 to 350, the normal approximation is appropriate.
P( x 100) P( z 0.60) .5 .2257 .2743 (Using Table IV, Appendix B)
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221
222
4.125
Chapter 4
a.
x is a binomial random variable with n = 1,000 and p = .28.
µ = E(x) = np = 1000(.28) = 280
b.
σ npq 1000(.28)(.72) 201.6 14.2
μ 3σ 280 3(14.2) 280 42.6 (237.4, 322.6)
Since the interval lies in the range 0 to 1000, we can use the normal approximation to approximate
the probability.
750 .5 280
P( x 750) P z
P ( z 33.1) .5 .5 0
14.2
4.126
a.
For n = 100 and p = .01:
μ ± 3σ np ± 3 npq 100(.01) ± 3 100(.01)(.99)
1 ± 3(.995) 1 ± 2.985 (−1.985, 3.985)
Since the interval does not lie in the range 0 to 100, we cannot use the normal approximation to
approximate the probabilities.
b.
For n = 100 and p = .5:
μ ± 3σ np ± 3 npq 100(.5) ± 3 100(.5)(.5) 50 ± 3(5) 50 ± 15 (35, 65)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probabilities.
c.
For n = 100 and p = .9:
μ ± 3σ np ± 3 npq 100(.9) ± 3 100(.9)(.1) 90 ± 3(3) 90 ± 9 (81, 99)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probabilities.
4.127
Let x = number of bottles (brands) selected in 65 trials that contain tap water. Then x is binomial random
variable with n = 65 and p = .25.
E(x) = μ = np = 65(.25) = 16.25
σ σ 2 npq 65(.25)(.75) 12.1875 3.49
To see if the normal approximation is appropriate, we use:
μ ± 3σ => 16.25 ± 3(3.49) => 16.25 ± 10.47 => (5.78, 26.72)
Since this interval lies in the range from 0 to 65, the normal approximation is appropriate.
(20 .5) 16.25
P( x 20) P z
P ( z .93) .5 P (0 z .93) .5 .3238 .1762
3.49
(Using Table IV, Appendix B)
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Random Variables and Probability Distributions
4.128
b.
223
Let v = number of credit card users out of 100 who carry Visa. Then v is a binomial random variable
with n = 100 and pv = .46.
E(v) = npv = 100(.46) = 46.
Let d = number of credit card users out of 100 who carry Discover. Then d is a binomial random
variable with n = 100 and pd = .06.
E(d) = npd = 100(.06) = 6.
c.
To see if the normal approximation is valid, we use:
μ 3σ npv 3 npv qv 100(.46) 3 100(.46)(.54) 46 3(4.984)
46 14.952 (31.048, 60.952)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probability.
(50 .5) 46
P(v 50) P z
P( z .70) .5 .2580 .2420
4.984
Let a = number of credit card users out of 100 who carry American Express. Then a is a binomial
random variable with n = 100 and pa = .12. To see if the normal approximation is valid, we use:
μ 3σ npa 3 npa qa 100(.12) 3 100(.12)(.88) 12 3(3.250)
12 9.75 (2.25, 21.75)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probability.
(50 .5) 12
P(a 50) P z
P( z 11.54) .5 .5 0
3.25
d.
4.129
In order for the normal approximation to be valid, μ ± 3σ must lie in the interval (0, n). This check
was done in part c for both portions of the question. In both cases, the normal approximation was
justified.
Let x = number of defective CDs in n = 1,600 trials. Then x is a binomial random variable with
n = 1,600 and p = .006.
E(x) = μ = np = 1,600(.006) = 9.6.
σ σ 2 npq 1, 600(.006)(.994) 9.5424 3.089
To see if the normal approximation is appropriate, we use:
μ 3σ 9.6 3(3.089) 9.6 9.267 (0.333, 18.867)
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224
Chapter 4
Since the interval lies in the range of 0 to 1,600, the normal approximation is appropriate.
11.5 9.6
P( x 12) P z
P ( z 0.62) .5 .2324 .2676
3.089
(Using Table IV, Appendix B)
Since this probability is fairly large, it would not be unusual to see 12 or more defectives in a sample of
1,600 if 99.4% were defect-free. Thus, there would be no evidence to cast doubt on the manufacturer’s
claim.
4.130
a.
There are well over a million college students in 4-year public and private institutions. In order to
collect a truly random sample, each of the college students must have an equal chance of being
selected. This would be extremely hard to do.
b.
Suppose your institution was a 4-year public institution. Let x = number of students receiving
financial aid in 100 trials. The random variable x has a binomial distribution with n = 100 and p =
.45. To determine if the normal approximation is appropriate, we check:
μ ± 3σ np ± 3 npq 100(.45) ± 3 100(.45)(.55)
45 ± 3(4.9749) 45 ± 14.9247 (30.0753, 59.9247)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probabilities.
(50 .5) 45
P(x ≥ 50) ≈ P z
= P(z ≥ .90) = .5 − .3159 = .1841
4.9749
(25 .5) 45
P(x < 25) ≈ P z
= P(z < −4.12) = .5 − .5 = 0
4.9749
Suppose your institution was a 4-year private institution. Let x = number of students receiving
financial aid in 100 trials. The random variable x has a binomial distribution with n = 100 and p =
.52. To determine if the normal approximation is appropriate, we check:
μ ± 3σ np ± 3 npq 100(.52) ± 3 100(.52)(.48)
52 ± 3(4.996) 52 ± 14.988 (37.012, 66.988)
Since the interval lies in the range 0 to 100, we can use the normal approximation to approximate the
probabilities.
(50 .5) 52
P(x ≥ 50) ≈ P z
= P(z ≥ −.50) = .5 + .1915 = .6915
4.996
(25 .5) 52
P(x < 25) ≈ P z
= P(z < −5.50) ≈ .5 − .5 = 0
4.996
c.
In order for the normal approximation to be appropriate, the interval μ ± 3σ should lie in the interval
0 to n. This assumption was checked in part b.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.131
225
Let x = number patients out of 150 who wait more than 30 minutes to see a doctor in a typical
U.S. emergency room. Then x is a binomial random variable with n = 150 and p = .5.
µ = np = 150(.5) = 75
σ npq 150(.5)(.5) 37.5 6.124
To see if the normal approximation is valid, we use:
μ 3σ 75 3(6.124) 75 18.372 (56.628, 93.372)
Since the interval lies in the range 0 to 150, we can use the normal approximation to approximate the
probability.
a.
b.
75 .5 75
P( x 75) P z
P( z .08) .5 .0319 .4681
6.124
(Using Table IV, Appendix B)
85 .5 75
P( x 85) P z
P( z 1.71) .5 .4564 .0436
6.124
(Using Table IV, Appendix B)
c.
90 .5 75
60 .5 75
P(60 x 90) P
z
P(2.37 z 2.37)
6.124
6.124
P(2.37 z 0) P(0 z 2.37) .4911 .4911 .9822
(Using Table IV, Appendix B)
4.132
a.
Let x = number of passengers in 1500 who will be detained for luggage inspection. Then x is a
binomial random variable with n = 1500 and p = .20. The expected number of passengers detained
will be:
E(x) = np = 1,500(.2) = 300
4.133
b.
For n = 4,000, E(x) = np = 4,000(.2) = 800
c.
(600 .5) 800
P(x > 600) ≈ P z
= P(z > −7.89) = .5 + .5 = 1.0
4000(.2)(.8)
a.
f(x) =
1
(c ≤ x ≤ d)
d c
1
1
1
= .04
d c 45 20 25
.04 (20 x 45)
So, f(x) =
0 otherwise
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226
Chapter 4
b.
c d 20 45 65
= 32.5
2
2
2
d c 45 20
= 7.22
σ=
12
12
μ=
c.
μ ± 2σ 32.5 ± 2(7.22) (18.06, 46.94)
P(18.06 < x < 46.94) = P(20 < x < 45) = (45 − 20).04 = 1
4.134
.04 (20 x 45)
From Exercise 4.133, f(x) =
0 otherwise
a.
P(20 ≤ x ≤ 30) = (30 − 20)(.04) = .4
b.
P(20 < x < 30) = (30 − 20)(.04) = .4
c.
P(x ≥ 30) = (45 − 30)(.04) = .6
d.
P(x ≥ 45) = (45 − 45)(.04) = 0
e.
P(x ≤ 40) = (40 − 20)(.04) = .8
f.
P(x < 40) = (40 − 20)(.04) = .8
g.
P(15 ≤ x ≤ 35) = (35 − 20)(.04) = .6
h.
P(21.5 ≤ x ≤ 31.5) = (31.5 − 21.5)(.04) = .4
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.135
a.
1
(c ≤ x ≤ d)
d c
1
1
1
d c 73 4
1
(3 x 7)
f(x) = 4
0 otherwise
f(x) =
c d 3 7 10
=5
2
2
2
d c 73
4
= 1.155
σ=
12
12
12
b.
μ=
c.
μ ± σ 5 ± 1.155 (3.845, 6.155)
P(μ − σ ≤x ≤ μ + σ) = P(3.845 ≤ x ≤ 6.155) =
4.136
4.137
4.138
4.139
227
a.
If θ = 1, a = 1, then e−a/θ = e−1 = .367879.
b.
If θ = 1, a = 2.5, then e−a/θ = e−2.5 = .082085
c.
If θ = .4, a = 3, then e−a/θ = e−7.5 = .000553
d.
If θ = .2, a = .3, then e−a/θ = e−1.5 = .223130
b a 6.155 3.845
2.31
=
d c
73
4
= .5775
P(x ≥ a) = e−a/θ = e−a/1. Using Table V, Appendix B:
a.
P(x > 1) = e−1/1 = e−1 = .367879
b.
P(x ≤ 3) = 1 − P(x > 3) = 1 − e−3/1 = 1 − e−3 = 1 − .049787 = .950213
c.
P(x > 1.5) = e−1.5/1 = e−1.5 = .223130
d.
P(x ≤ 5) = 1 − P(x > 5) = 1 − e−5/1 = 1 − e−5 = 1 − .006738 = .993262
a.
P(x ≤ 4) = 1 − P(x > 4) = 1 − e−4/2.5 = 1 − e−1.6 = 1 − .201897 = .798103 (Using Table V, Appendix B)
b.
P(x > 5) = e−5/2.5 = e−2 = .135335
c.
P(x ≤ 2) = 1 − P(x > 2) = 1 − e−2/2.5 = 1 − e −.8 = 1 − .449329 = .550671
d.
P(x > 3) = e−3/2.5 = e−1.2 = .301194
1
1
1
= .01
d c 200 100 100
.01 (100 x 200)
f(x) =
0 otherwise
f(x) =
c d 100 200 300
= 150
2
2
2
d c 200 100 100
σ=
= 28.8675
12
12
12
μ=
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228
Chapter 4
a.
μ ± 2σ 150 ± 2(28.8675) 150 ± 57.735 (92.265, 207.735)
P(x < 92.265) + P(x > 207.735) = P(x < 100) + P(x > 200)
=
0
+
0
=0
b.
μ ± 3σ 150 ± 3(28.8675) 150 ± 86.6025 (63.3975, 236.6025)
P(63.3975 < x < 236.6025) = P(100 < x < 200) = (200 − 100)(.01) = 1
c.
From a, μ ± 2σ (92.265, 207.735).
P(92.265 < x < 207.735) = P(100 < x < 200) = (200 − 100)(.01) = 1
4.140
With θ = 2, f(x)=
1 −x/2
(x > 0)
e
2
μ=σ=θ=2
a.
μ ± 3σ 2 ± 3(2) (−4, 8)
Since μ − 3σ lies below 0, find the probability that x is more than μ + 3σ = 8.
P(x > 8) = e−8/2 = e−4 = .018316
b.
(using Table V, Appendix B)
μ ± 2σ 2 ± 2(2) (−2, 6)
Since μ ± 2σ lies below 0, find the probability that x is between 0 and 6.
P(x < 6) = 1 − P(x ≥ 6) = 1 − e−6/2 = 1 − e−3 = 1 − .049787 = .950213
(using Table V, Appendix B)
c.
μ ± .5σ 2 ± .5(2) (1, 3)
P(1 < x < 3) = P(x > 1) − P(x > 3)
= e−1/2 − e−3/2 = e−.5 − e−1.5
= .606531 − .223130
= .383401 (using Table V in Appendix B)
4.141
a.
Let x = temperature with no bolt-on trace elements. Then x has a uniform distribution.
f ( x)
1
d c
(c x d )
1
1
1
d c 290 260 30
1
Therefore, f ( x) 30
0
(260 x 290)
otherwise
P(280 x 284) (284 280)
1
1
4 .133
30
30
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Random Variables and Probability Distributions
229
Let y = temperature with bolt-on trace elements. Then y has a uniform distribution.
f ( y)
1
d c
(c y d )
1
1
1
d c 285 278 7
1
(278 y 285)
Therefore, f ( y ) 7
0
otherwise
P(280 y 284) (284 280)
b.
P( x 268) (268 260)
1
1
4 .571
7
7
1
1
8 .267
30
30
P( y 268) (268 260)(0) 0
4.142 a.
Let x = number of anthrax spores. Then x has an approximate uniform distribution.
f ( x)
1
d c
(c ≤ x ≤ d)
1
1
1
.1
d c 10 0 10
.1
Therefore, f ( x)
0
(0 x 10)
otherwise
P(x ≤ 8) = (8 – 0)(.1) = .8
b.
4.143
4.144
P(2 ≤ x ≤ 5) = (5 – 2)(.1) = .3
a.
P(x > 2) = e−2/2.5 = e−.8 = .449329
b.
P(x < 5) = 1 – P(x ≥ 5) = 1 – e−5/2.5 = 1 – e−2 = 1 − .135335 = .864665
(using Table V, Appendix B)
a.
Let x = time until the first critical part failure. Then x has an exponential distribution with θ = .1.
(using Table V, Appendix B)
P( x 1) e 1/.1 e 10 .0000454 (using Table V, Appendix B)
b.
30 minutes = .5 hours. P( x .5) 1 P ( x .5) 1 e .5/.1 1 e5 1 .0067 .9933
(using table V, Appendix B)
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230
4.145
Chapter 4
To construct a relative frequency histogram for the data, we can use 7 measurement classes.
Interval width =
Largest number - smallest number
98.0716 .7434
=
= 13.9
Number of classes
7
We will use an interval width of 14 and a starting value of .74335.
The measurement classes, frequencies, and relative frequencies are given in the table below.
Class
Measurement Class
Class Frequency
1
2
3
4
5
6
7
.74335 − 14.74335
14.74335 − 28.74335
28.74335 − 42.74335
42.74335 − 56.74335
56.74335 − 70.74335
70.74335 − 84.74335
84.74335 − 98.74335
6
4
6
6
5
4
9
40
Class Relative
Frequency
6/40 = .15
.10
.15
.15
.125
.10
.225
1.000
The histogram looks like the data could
be from a uniform distribution. The last
class (84.74335 − 98.74335) has a few
more observations in it than we would
expect. However, we cannot expect a
perfect graph from a sample of only 40
observations.
4.146
a.
For layer 2, let x = amount loss. Since the amount of loss is random between .01 and .05 million
dollars, the uniform distribution for x is:
f(x) =
1
d c
(c ≤ x ≤ d)
1
1
1
= 25
d c .05 .01 .04
25 (.01 x .05)
Therefore, f(x) =
0 otherwise
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
A graph of the distribution looks like the following:
μ=
σ=
c+d
.01 + .05
=
= .03
2
2
d c
12
.05 .01
12
= .0115, σ2 = (.0115)2 = .00013
The mean loss for layer 2 is .03 million dollars and the variance of the loss for layer 2 is .00013
million dollars squared.
b.
For layer 6, let x = amount loss. Since the amount of loss is random between .50 and 1.00 million
dollars, the uniform distribution for x is:
f(x) =
1
d c
(c ≤ x ≤ d )
1
1
1
=2
d c 1.00 .50 .50
2 (.50 x 1.00)
Therefore, f(x) =
0 otherwise
A graph of the distribution looks like the following:
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231
232
Chapter 4
μ=
σ=
c d
.50 1.00
= .75
2
2
d c
12
1.00 .50
12
= .1443, σ2 = (.1443)2 = .0208
The mean loss for layer 6 is .75 million dollars and the variance of the loss for layer 6 is .0208
million dollars squared.
c.
A loss of $10,000 corresponds to x = .01. P(x > .01) = 1
A loss of $25,000 corresponds to x = .025.
1
1
P(x < .025) = (Base)(Height) = (x − c)
= (.025 − .01)
d c
.05 .01
d.
= .015(25) = .375
A loss of $750,000 corresponds to x = .75. A loss of $1,000,000 corresponds to x = 1.
1
1
P(.75 < x < 1) = (Base)(Height) = (d - x)
= (1.00 - .75)
d c
1.00 .50
= .25(2) = .5
A loss of $900,000 corresponds to x = .90.
1
1
P(x > .9) = (Base)(Height) = (d − x)
= (1.00 − .90)
d c
1.00 .50
= .10(2) = .20
P(x = .9) = 0
4.147
a.
The amount dispensed by the beverage machine is a continuous random variable since it can take on
any value between 6.5 and 7.5 ounces.
b.
Since the amount dispensed is random between 6.5 and 7.5 ounces, x is a uniform random variable.
f ( x)
1
(c ≤ x ≤ d)
d c
1
1
1
=1
d c 7.5 6.5 1
1 (6.5 x 7.5)
Therefore, f ( x)
0 otherwise
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
The graph is as follows:
c.
μ
σ
c d 6.5 7.5 14
7
2
2
2
d c
12
7.5 6.5
12
.2887
μ ± 2σ 7 ± 2(.2887) 7 ± .5774 (6.422, 7.577)
d.
P(x ≥ 7) = (7.5 − 7)(1) = .5
e.
P(x < 6) = 0
f.
P(6.5 ≤ x ≤ 7.25) = (7.25 − 6.5)(1) = .75
g.
The probability that the next bottle filled will contain more than 7.25 ounces is:
P(x > 7.25) = (7.5 − 7.25)(1) = .25
The probability that the next 6 bottles filled will contain more than 7.25 ounces is:
P[(x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25) ∩ (x > 7.25)]
= [P(x >7.25)]6 = .256 = .0002
4.148
a.
Let x = length of time elapsed before the winning goal is scored. Then x has an exponential
distribution with μ = 9.15.
P(x ≤ 3) = 1 − P(x > 3) = 1 − e−3/9.15 = 1 − e−.327869
= 1 − .720457 = .279543
b.
P(x > 20) = e−20/9.15 = e−2.185792 = .1123887
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233
234
a.
For μ = 17 = θ. To graph the distribution, we will pick several values of x and find the value of f(x),
where x = time between arrivals of the smaller craft at the pier.
f(x) =
1
θ
e x /θ
1 x /17
e
17
1 1/17
= .0555
e
17
1
f(3) = e3/17 = .0493
17
1 5 /17
= .0438
f(5) =
e
17
1 7 /17
= .0390
f(7) =
e
17
1 10 /17
= .0327
f(10) =
e
17
1 15/17
= .0243
f(15) =
e
17
1 20 /17
= .0181
f(20) =
e
17
1 25/17
= .0135
f(25) =
e
17
f(1) =
The graph is:
0.06
0.05
0.04
f(x)
4.149
Chapter 4
0.03
0.02
0.01
0
5
10
15
20
25
x
b.
We want to find the probability that the time between arrivals is less than 15 minutes.
P(x < 15) = 1 − P(x ≥ 15) = 1 − e−15/17 = 1 − .4138 = .5862
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.150
Let x = cycle availability, where x has a uniform distribution on the interval from 0 to 1.
Mean = μ
c d 0 1
.5
2
2
Standard deviation = σ
d c
12
1 0
12
.289
The 10th percentile is that value of x such that 10% of all observations are below it.
Let K1 = 10th percentile.
P(x ≤ K1) = (K1 − 0)(1 − 0) = K1 = .10
The lower quartile is that value of x such that 25% of all observations are below it.
Let K2 = 25th percentile.
P(x ≤ K2) = (K2 − 0)(1 − 0) = K2 = .25
The UPPER quartile is that value of x such that 75% of all observations are below it.
Let K3 = 75th percentile.
P(x ≤ K3) = (K3 − 0)(1 − 0) = K3 = .75
4.151
a.
Let x = product’s lifetime at the end of its lifetime. Then x has an exponential distribution
with μ = 500,000.
P(x < 700,000) = 1 – P(x ≥ 700,000) = 1 – e-700000/5000000 = 1 – e-1.4 = 1 − .246597 = .753403
(Using Table V, Appendix B)
b.
Let y = product’s lifetime during its normal life. Then y has a uniform distribution.
f ( y)
1
d c
(c ≤ y ≤ d)
1
1
1
d c 1, 000, 000 100, 000 900, 000
1
Therefore, f ( y ) 900,000
0
(100,000 y 1,000,000)
otherwise
1
P( y 700, 000) (700, 000 100, 000)
.667
900, 000
c.
P(x < 830,000) = 1 – P(x ≥ 830,000) = 1 – e-830000/5000000 = 1 – e-1.66 = 1 − .190139 = .809861
(Using a calculator)
1
P( y 830, 000) (830, 000 100, 000)
.811
900, 000
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235
236
4.152
Chapter 4
a.
Let x = life length of CD-ROM. Then x has an exponential distribution with θ = 25,000.
R(t ) P ( x t ) e t / 25,000
b.
R(8,760) P ( x 8, 760) e8,760 / 25,000 e.3504 .7044
c.
S(t) = probability that at least one of two drives has a length exceeding t hours
= 1 – probability that neither has a length exceeding t hours
= 1 – P(x1 ≤ t)P(x2 ≤ t) = 1 – [1 – P(x1 > t)][1 – P(x2> t)]
= 1 – [1 – e-t/25,000][1 – e-t/25,000]
=1 – [1 – 2e-t/25,000 + e-t/12,500] = 2e-t/25,000 – e-t/12,500
4.153
d.
S (8, 760) 2e 8,760 / 25,000 e8,760 /12,500 2(.7044) .4962 1.4088 .4962 .9126
e.
The probability in part d is greater than that in part b. We would expect this. The probability that at
least one of the systems lasts longer than 8,760 hours would be greater than the probability that only
one system lasts longer than 8,760 hours.
Let x = number of inches a gouge is from one end of the spindle. Then x has a uniform distribution with
f(x) as follows:
1
1
1
f(x) = d c 18 0 18
0
0 x 18
otherwise
In order to get at least 14 consecutive inches without a gouge, the gouge must be within 4 inches of either
end. Thus, we must find:
P(x < 4) + P(x > 14) = (4 − 0)(1/18) + (18 − 14)(1/18) = 4/18 + 4/18 = 8/18 = .4444
4.154
a.
b.
c.
c d 0 1
= .5
2
2
d c 1 0
= .289
σ=
12
12
μ=
σ2 = .2892 = .083
P(p > .95) = (1 − .95)(1) = .05
P(p < .95) = (.95 − 0)(1) = .95
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
d.
237
The analyst should use a uniform probability distribution with c = .90 and d = .95.
1
1
1
=
=
= 20 (.90 ≤ p ≤ .95)
f(p) = d − c .95 − .90 .05
0 otherwise
4.155
a.
For θ = 250, P(x > a) = e−a/250
For a = 300 and b = 200, show P(x > a + b) ≥ P(x > a)P(x > b)
P(x > 300 + 200) = P(x > 500) = e−500/250 = e−2 = .1353
P(x > 300) P(x > 200) = e−300/250 e−200/250 = e−1.2 e−.8 = .3012(.4493) = .1353
Since P(x > 300 + 200) = P(x > 300) P(x > 200), then
P(x > 300 + 200) ≥ P(x > 300) P(x > 200)
Also, show P(x > 300 + 200) ≤P(x > 300) P(x > 200). Since we already showed that
P(x > 300 + 200) = P(x > 300) P(x > 200),
then P(x > 300 + 200) ≤ P(x > 300) P(x > 200).
b.
Let a = 50 and b = 100. Show P(x > a + b) ≥ P(x > a) P(x > b)
P(x > 50 + 100) = P(x > 150) = e−150/250 = e−.6 = .5488
P(x > 50) P(x > 100) = e−50/250 e−100/250 = e−.2e−.4 = .8187(.6703) = .5488
Since P(x > 50 + 100) = P(x > 50) P(x > 100), then
P(x > 50 + 100) ≥ P(x > 50) P(x > 100)
Also, show P(x > 50 + 100) ≤ P(x > 50) P(x > 100). Since we already showed that
P(x > 50 + 100) = P(x > 50) P(x > 100),
then P(x > 50 + 100) ≤ P(x > 50) P(x > 100).
c.
4.156
Show P(x > a + b) ≥ P(x > a) P(x > b)
P(x > a + b) = e−(a+b)/250 = e−a/250 e−b/250 = P(x > a) P(x > b)
a–b. The different samples of n = 2 with replacement and their means are:
Possible Samples
0, 0
0, 2
0, 4
0, 6
2, 0
2, 2
2, 4
2, 6
c.
x
0
1
2
3
1
2
3
4
Possible Samples
4, 0
4, 2
4, 4
4, 6
6, 0
6, 2
6, 4
6, 6
x
2
3
4
5
3
4
5
6
Since each sample is equally likely, the probability of any 1 being selected is
1 1 1
4 4 16
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238
Chapter 4
d.
P( x = 0) =
P( x = 1) =
P( x = 2) =
P( x = 3) =
P( x = 4) =
P( x = 5) =
P( x = 6) =
1
16
1
1
2
16 16 16
1
1
1
3
16 16 16 16
1
1
1
1
4
16 16 16 16 16
1
1
1
3
16 16 16 16
1
1
2
16 16 16
1
16
x
p( x )
0
1
2
3
4
5
6
1/16
2/16
3/16
4/16
3/16
2/16
1/16
e.
4.157
Solution will vary. See page 1037 for Guided Solutions.
4.158
If the observations are independent of each other, then
P(1, 1) = p(1)p(1) = .2(.2) = .04
P(1, 2) = p(1)p(2) = .2(.3) = .06
P(1, 3) = p(1)p(3) = .2(.2) = .04
etc.
a.
Possible Sample
1, 1
1, 2
1, 3
1, 4
1, 5
2, 1
2, 2
2, 3
2, 4
2, 5
3, 1
3, 2
3, 3
x
1
1.5
2
2.5
3
1.5
2
2.5
3
3.5
2
2.5
3
p( x )
Possible Samples
.04
.06
.04
.04
.02
.06
.09
.06
.06
.03
.04
.06
.04
3, 4
3, 5
4, 1
4, 2
4, 3
4, 4
4, 5
5, 1
5, 2
5, 3
5, 4
5, 5
x
3.5
4
2.5
3
3.5
4
4.5
3
3.5
4
4.5
5
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
p( x )
.04
.02
.04
.06
.04
.04
.02
.02
.03
.02
.02
.01
Random Variables and Probability Distributions
Summing the probabilities, the probability distribution of is:
x
p( x )
1
1.5
2
2.5
3
3.5
4
4.5
5
.04
.12
.17
.20
.20
.14
.08
.04
.01
b.
4.159
c.
P( x ≥ 4.5) = .04 + .01 = .05
d.
No. The probability of observing = 4.5 or larger is small (.05).
E ( x) = μ =
xp( x) = 1(.2) + 2(.3) + 3(.2) + 4(.2) + 5(.1)
= .2 + .6 + .6 + .8 + .5 = 2.7
E( x ) =
xp( x ) = 1.0(.04) + 1.5(.12) + 2.0(.17) + 2.5(.20) + 3.0(.20) + 3.5(.14) + 4.0(.08)
+ 4.5(.04) + 5.0(.01)
= .04 + .18 + .34 + .50 + .60 + .49 + .32 + .18 + .05 = 2.7
4.160
a.
For a sample of size n = 2, the sample mean and sample median are exactly the same. Thus, the
sampling distribution of the sample median is the same as that for the sample mean (see Exercise
4.158a).
b.
The probability histogram for the sample median is identical to that for the sample mean (see
Exercise 4.158b).
4.161
Solution will vary. See page 1037 for Guided Solutions.
4.162
Solution will vary. See page 1037 for Guided Solutions.
4.163
The sampling distribution is approximately normal only if the sample size is sufficiently large or if the
population being sampled from is normal.
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239
240
4.164
4.165
4.166
Chapter 4
a.
μ x = μ = 10, σ x = σ / n 3 / 25 = 0.6
b.
μ x = μ = 100, σ x = σ / n 25 / 25 = 5
c.
μ x = μ = 20, σ x = σ / n 40 / 25 = 8
d.
μ x = μ = 10, σ x = σ / n 100 / 25 = 20
a.
μ x μ 100, σ x
b.
μ x μ 100, σ x
c.
μ x μ 100, σ x
d.
μ x μ 100, σ x
e.
μ x μ 100, σ x
f.
μ x μ 100, σ x
a.
μ x μ 20, σ x σ / n 16 / 64 2
b.
By the Central Limit Theorem, the distribution of is approximately normal. In order for
the Central Limit Theorem to apply, n must be sufficiently large. For this problem,
n = 64 is sufficiently large.
c.
z=
d.
z=
x μx
σx
x μx
σx
σ
n
σ
n
σ
n
σ
n
σ
n
σ
n
100
100
100
100
100
100
4
25
100
50
5
2
1
1.414
500
.447
1000
15.5 20
2.25
2
23 20
1.50
2
.316
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.167
In Exercise 4.166, it was determined that the mean and standard deviation of the sampling distribution of
the sample mean are 20 and 2 respectively. Using Table IV, Appendix B:
a.
16 20
P( x < 16) = P z
= P(z < −2) = .5 − .4772 = .0228
2
b.
23 20
P( x > 23) = P z
= P(z > 1.50) = .5 − .4332 = .0668
2
c.
25 20
P( x > 25) = P z
= P(z > 2.5) = .5 − .4938 = .0062
2
d.
e.
4.168
241
22 20
16 20
P(16 < x < 22) = P
z
= P(−2 < z < 1)
2
2
= .4772 + .3413 = .8185
14 20
P( x < 14) = P z
= P(z < −3) = .5 − .4987 = .0013
2
For this population and sample size,
E ( x ) = μ = 100, σ x = σ / n 10 / 900 = 1/3
a.
Approximately 95% of the time, will be within two standard deviations of the mean, i.e., μ ± 2σ
2
1
100 ± 2 100 ±
(99.33, 100.67). Almost all of the time, the sample mean will be within
3
3
1
three standard deviations of the mean, i.e., μ ± 3σ 100 ± 3 100 ± 1 (99, 101).
3
b.
c.
4.169
1
No more than three standard deviations, i.e., 3 = 1
3
No, the previous answer only depended on the standard deviation of the sampling distribution of the
sample mean, not the mean itself.
By the Central Limit Theorem, the sampling distribution of is approximately normal with μ x = μ = 30
and σ x σ / n 16 / 100 = 1.6. Using Table IV, Appendix B:
a.
28 30
P( ≥ 28) = P z
= P(z ≥ −1.25) = .5 + .3944 = .8944
1.6
b.
26.8 30
22.1 30
z
P(22.1 ≤ x ≤ 26.8) = P
= P(−4.94 ≤ z ≤ −2) = .5 − .4772 = .0228
1.6
1.6
c.
28.2 30
P( x ≤ 28.2) = P z
= P(z ≤ −1.13) = .5 − .3708 = .1292
1.6
d.
27.0 30
P( x ≥ 27.0) = P z
= P(z ≥ −1.88) = .5 + .4699 = .9699
1.6
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242
Chapter 4
4.170
Solution will vary. See page 1037 for Guided Solutions.
4.171
a.
μ x μ 141
b.
σx
c.
By the Central Limit Theorem, the sampling distribution of x is approximately normal.
d.
z
e.
P( x 142) P( z 0.56) .5 .2123 .2877 (Using Table IV, Appendix B)
a.
μ x μ 97,300
b.
σx
4.172
c.
4.173
σ
n
x μx
σx
σ
n
18
100
1.8
142 141
0.56
1.8
30, 000
50
4, 242.6407
By the Central Limit Theorem, the sampling distribution of x is approximately normal.
x μx
z
e.
P( x 89,500) P ( z 1.84) .5 .4671 .9671 (Using Table IV, Appendix B)
σx
89,500 97,300
1.84
4, 242.6407
d.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with μ x μ 19
and σ x
σ
n
65
100
6.5 .
Using Table IV, Appendix B,
10 19
P( x 10) P z
P ( z 1.38) .5 .4162 .0838
6.5
4.174
a.
b.
c.
4.175
a.
b.
σ
.10
.0141
n
50
Since n > 30, the sampling distribution of x is approximately normal by the Central Limit Theorem.
E ( x ) μ x μ .10
σx
.13 .10
P( x .13) P z
P ( z 2.13) .5 .4834 .0166
.0141
(Using Table IV, Appendix B)
By the Central Limit Theorem, the sampling distribution of x is approximately normal with a mean
σ
.193
.0273 .
μ x μ .53 and standard deviation σ x
n
50
.58 .53
P( x .58) P z
P ( z 1.83) .5 .4664 .0336
.0273
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Random Variables and Probability Distributions
c.
243
If Before Tensioning: μ x μ .53
.59 .53
P( x .59) P z
P( z 2.20) .5 .4861 .0139
.0273
If After Tensioning: μ x μ .58
.59 .58
P( x .59) P z
P( z 0.37) .5 .1443 .3557
.0273
Since the probability of getting a maximum differential of .59 or more Before Tensioning is so small,
it would be very unlikely that the measurements were obtained before tensioning. However, since
the probability of getting a maximum differential of .59 or more After Tensioning is not small, it
would not be unusual that the measurements were obtained after tensioning. Thus, most likely, the
measurements were obtained After Tensioning.
4.176
a.
Since the sample size is small, we also have to assume that the distribution from which the sample
σ
.5
was drawn is normal. μ x μ 1.8 , σ x
.1118
n
20
1.85 1.8
P( x 1.85) P z
P ( z 0.45) .5 .1736 .3264
.1118
(using Table IV, Appendix B)
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Rough
Variable
Rough
N
20
N*
0
Mean
1.881
SE Mean
0.117
StDev
0.524
Minimum
1.060
Q1
1.303
Median
2.040
Q3
2.293
Maximum
2.640
From this output, the value of x is 1.881.
c.
For x = 1.881:
1.881 1.8
P( x 1.881) P z
P( z 0.72) .5 .1736 .3264
.1118
Since this probability is so high, observing a sample mean of x = 1.881, is not unusual. The
assumptions in part a appear to be valid.
4.177
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with a mean
σ
10
μ x μ 6 and standard deviation σ x
.5538 .
n
326
7.5 6
P( x 7.5) P z
P( z 2.71) .5 .4966 .0034
.5538
(Using Table IV, Appendix B)
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244
Chapter 4
b.
We first need to find the probability of observing the current data or anything more unusual
if the true mean is 6.
300 6
P( x 300) P z
P( z 530.88) .5 .5 0
.5538
Since the probability of observing a sample mean of 300 ppb or higher is essentially 0 if the true
mean is 6 ppb, we would infer that the true mean PFOA concentration for the population of people
who live near DuPont’s Teflon facility is not 6 ppb but higher than 6 ppb.
4.178
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal with
μ x μ and σ x σ / n σ / 100 .
b.
The mean of the x distribution is equal to the mean of the distribution of the fleet or the
fleet mean score.
c.
μ x μ 30 and σ x σ / n σ / 100 60 / 100 6 .
d.
4.179
45 30
P( x 45) P z
P z 2.5 .5 .4938 .0062 (Using Table IV, Appendix B)
6
The sample mean of 45 tends to refute the claim. If the true fleet mean was as high as 30, observing
a sample mean of 45 or higher would be extremely unlikely (probability = .0062). Thus, we would
infer that the true mean is actually not 30 but something higher. Thus, we would refute the
company’s claim that the mean “couldn’t possibly be as large as 30.”
a.
By the Central Limit Theorem, the sampling distribution of is approximately normal
with μ x = μ and σ x = σ / n = σ / 50 .
b.
μ x = μ = 40 and σ x = σ / 50 = 12 / 50 = 1.6971.
44 40
P( x ≥ 44) = P z
= P(z ≥ 2.36) = .5 − .4909 = .0091
1.6971
(using Table IV, Appendix B)
c.
μ ± 2σ / n 40 ± 2(1.6971) 40 ± 3.3942 (36.6058, 43.3942)
43.3942 40
36.6058 40
P(36.6058 ≤ x ≤ 43.3942) = P
z
1.6971
1.6971
= P(−2 ≤ z ≤ 2) = 2(.4772) = .9544
(using Table IV, Appendix B)
4.180
For n = 36, μ x = μ = 406 and σ x = σ / n = 10.1/ 36 = 1.6833. By the Central Limit Theorem, the
sampling distribution is approximately normal (n is large).
400.8 406
P( x ≤ 400.8) = P z
= P(z ≤ −3.09) = .5 − .4990 = .0010
1.6833
(using Table IV, Appendix B)
The first. If the true value of μ is 406, it would be extremely unlikely to observe an as small as 400.8 or
smaller (probability .0010). Thus, we would infer that the true value of μ is less than 406.
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Random Variables and Probability Distributions
4.181
245
For n = 50, we can use the Central Limit Theorem to decide the shape of the distribution of the sample
mean bacterial counts. For the handrubbing sample, the sampling distribution of x is approximately
σ
59
8.344 . For the handwashing sample,
normal with a mean of μ = 35 and standard deviation
n
50
the sampling distribution of x is approximately normal with a mean of μ = 69 and standard deviation
σ
106
14.991 .
n
50
For Handrubbing:
30 35
P( x 30 | μ 35) P z
P ( z .60) .5 .2257 .2743
8.344
(using Table IV, Appendix B)
For Handwashing:
30 69
P( x 30 | μ 69) P z
P( z 2.60) .5 .4953 .0047
14.991
(using Table IV, Appendix B)
Since the probability of getting a sample mean of less than 30 for the handrubbing is not small compared
with that for the handwashing, the sample of workers probably came from the handrubbing group.
4.182
a.
This experiment consists of 100 trials. Each trial results in one of two outcomes: chip is defective or
not defective. If the number of chips produced in one hour is much larger than 100, then we can
assume the probability of a defective chip is the same on each trial and that the trials are independent.
Thus, x is a binomial. If, however, the number of chips produced in an hour is not much larger than
100, the trials would not be independent. Then x would not be a binomial random variable.
b.
This experiment consists of two trials. Each trial results in one of two outcomes: applicant qualified
or not qualified. However, the trials are not independent. The probability of selecting a qualified
applicant on the first trial is 3 out of 5. The probability of selecting a qualified applicant on the
second trial depends on what happened on the first trial. Thus, x is not a binomial random variable.
It is a hypergeometric random variable.
c.
The number of trials is not a specified number in this experiment, thus x is not a binomial random
variable. In this experiment, x is counting the number of calls received.
d.
The number of trials in this experiment is 1000. Each trial can result in one of two outcomes: favor
state income tax or not favor state income tax. Since 1000 is small compared to the number of
registered voters in Florida, the probability of selecting a voter in favor of the state income tax is the
same from trial to trial, and the trials are independent of each other. Thus, x is a binomial random
variable.
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246
4.183
4.184
Chapter 4
n
p(x) = p x q n - x x = 0, 1, 2, ... , n
x
a.
7
7! 3 4
P(x = 3) = p(3) = .53.5 4
.5 .5 = 35(.125)(.0625) = .2734
3!4!
3
b.
4
4! 3 1
P(x = 3) = p(3) = .83.21
.8 .2 = 4(.512)(.2) = .4096
3!1!
3
c.
15
15! 1 14
.1 .9 = 15(.1)(.228768) = .3432
P(x = 1) = p(1) = .11.914
1!14!
1
a.
μ=
σ2 =
xp( x) = 10(.2) + 12(.3) + 18(.1) + 20(.4) = 15.4
(x μ )
2
p( x)
= (10 − 15.4) (.2) + (12 − 15.4)2(.3) + (18 − 15.4)2(.1) + (20 − 15.4)2(.4) = 18.44
σ = 18.44 ≈ 4.294
4.185
2
b
P(x < 15) = p(10) + p(12) = .2 + .3 = .5
c.
μ ± 2σ = 15.4 ± 2(4.294) (6.812, 23.988)
d.
P(6.812 < x < 23.988) = .2 + .3 + .1 + .4 = 1.0
From Table II, Appendix B:
a.
P(x = 14) = P(x ≤ 14) − P(x ≤ 13) = .584 − .392 = .192
b.
P(x ≤ 12) = .228
c.
P(x > 12) = 1 − P(x ≤ 12) = 1 − .228 = .772
d.
P(9 ≤ x ≤ 18) = P(x ≤ 18) − P(x ≤ 8) = .992 − .005 = .987
e.
P(8 < x < 18) = P(x ≤ 17) − P(x ≤ 8) = .965 − .005 = .960
f.
μ = np = 20(.7) = 14
σ2 = npq = 20(.7)(.3) = 4.2, σ =
g.
4.2 = 2.049
μ ± 2σ 14 ± 2(2.049) 14 ± 4.098 (9.902, 18.098)
P(9.902 < x < 18.098) = P(10 ≤ x ≤ 18) = P(x ≤ 18) − P(x ≤ 9) = .992 − .017 = .975
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.186
4.187
4.188
4.189
4.190
Using Table III, Appendix B,
a.
When λ = 2, p(3) = P(x ≤ 3) − P(x ≤ 2) = .857 − .677 = .180
b.
When λ = 1, p(4) = P(x ≤ 4) − P(x ≤ 3) = .996 − .981 = .015
c.
When λ = .5, p(2) = P(x ≤ 2) − P(x ≤ 1) = .986 − .910 = .076
a.
Poisson
b.
Binomial
c.
Binomial
a.
r N r 3 8 3
3! 5!
x n x 2 5 2
2!1!
3!2! 3(10) = .536
P(x = 2) =
8!
56
N
8
n
5
5!3!
b.
r N r 2 6 2
2! 4!
x n x 2 2 2
1(1)
2!0! 0!4!
= .067
P(x = 2) =
6!
15
N
6
n
2
2!4!
c.
r N r 4 5 4
4! 1!
x n x 3 4 3
3!1!
1!0! 4(1) = .8
P(x = 3) =
5!
5
N
5
n
3
4!1!
a.
Discrete - The number of damaged inventory items is countable.
b.
Continuous - The average monthly sales can take on any value within an acceptable limit.
c.
Continuous - The number of square feet can take on any positive value.
d.
Continuous - The length of time we must wait can take on any positive value.
a.
1
1
1
,10 x 90
f(x) = d c 90 10 80
0
otherwise
b.
μ=
cd
10 90
=
= 50
2
2
d c
90 10
=
= 23.094011
σ=
12
12
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
247
248
4.191
Chapter 4
c.
The interval μ ± 2σ 50 ± 2(23.094)
50 ± 46.188 (3.812, 96.188) is indicated on the
graph.
d.
P(x ≤ 60) = Base(height) = (60 − 10)
e.
P(x ≥ 90) = 0
f.
P(x ≤ 80) = Base(height) = (80 − 10)
g.
P(μ −σ ≤ x ≤ μ + σ) = P(50 − 23.094 ≤ x ≤ 50 + 23.094)
= P(26.906 ≤ x ≤ 73.094)
= Base(height)
1 46.188
= (73.094 − 26.906)
= .577
80
80
h.
P(x > 75) = Base(height) = (90 − 75)
a.
P(z ≤ 2.1) = A1 + A2
= .5 + .4821
= .9821
b.
P(z ≥ 2.1) = A2 = .5 − A1
= .5 − .4821
= .0179
c.
P(z ≥ −1.65) = A1 + A2
= .4505 + .5000
= .9505
d.
P(−2.13 ≤ z ≤ −.41)
= P(−2.13 ≤ z ≤ 0) − P(−.41 ≤ z ≤ 0)
= .4834 − .1591
= .3243
1 5
= .625
80 8
1 7
= .875
80 8
1 15
= .1875
80 80
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.192
249
e.
P(−1.45 ≤ z ≤ 2.15) = A1 + A2
= .4265 + .4842
= .9107
f.
P(z ≤ −1.43) = A1 = .5 − A2
= .5000 − .4236
= .0764
a.
P(z ≤ z0) = .5080
P(0 ≤ z ≤ z0) = .5080 − .5 = .0080
Looking up the area .0080 in Table IV,
z0 = .02
b.
P(z ≥ z0) = .5517
P(z0 ≤ z ≤ 0) = .5517 − .5 = .0517
Looking up the area .0517 in Table IV, z0 = −.13.
c.
P(z ≥ z0) = .1492
P(0 ≤ z ≤ z0) = .5 − .1492 = .3508
Looking up the area .3508 in Table IV,
z0 = 1.04
d.
P(z0 ≤ z ≤ .59) = .4773
P(z0 ≤ z ≤ 0) + P(0 ≤ z ≤ .59) = .4773
P(0 ≤ z ≤ .59) = .2224
Thus, P(z0 ≤ z ≤ 0) = .4773 − .2224 = .2549
Looking up the area .2549 in Table IV, z0 = -.69
4.193
x
7
a.
For the probability density function, f ( x)
e
b.
For the probability density function, f ( x)
1
, 5 x 25 , x is a uniform random variable.
20
c.
For the probability function, f ( x)
7
e .5[( x 10) / 5]
, x 0 , x is an exponential random variable.
2
5 2π
, x is a normal random variable.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
250
4.194
4.195
Chapter 4
a.
P(x ≤ 1) = 1 − P(x > 1) = 1 − e−1/3 = 1 − .716531 = .283469 (using calculator)
b.
P(x > 1) = e −1/3 = .716531
c.
P(x = 1) = 0
d.
P(x ≤ 6) = 1 − P(x > 6) = 1 − e−6/3 = 1 − e−2 = 1 − .135335 = .864665 (using Table V, Appendix B)
e.
P(2 ≤ x ≤ 10) = P(x ≥ 2) − P(x > 10) = e−2/3 − e−10/3
= .513417 − .035674
= .477743 (using calculator)
a.
b.
c.
(x is a continuous random variable. There is no probability associated with a single
point.)
80 75
P(x ≤ 80) = P z
= P(z ≤ .5)
10
= .5000 + .1915 = .6915
(Table IV, Appendix B)
85 75
P(x ≥ 85) = P z
= P(z ≥ 1)
10
= .5000 - .3413 = .1587
(Table IV, Appendix B)
75 75
70 75
z
P(70 ≤ x ≤ 75) = P
10
10
= P(−.5 ≤ z ≤ 0)
= P(0 ≤ z ≤ .5) = .1915
d.
P(x > 80) = 1 − P(x ≤ 80) = 1 − .6915 = .3085 (Refer to part a.)
e.
P(x = 78) = 0, since a single point does not have an area.
f.
110 75
P(x ≤ 110) = P z
= P(z ≤ 3.5)
10
≈ .5000 + .5000 = 1.0
(Table IV, Appendix B)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.196
4.174 μ = np = 100(.5) = 50, σ =
a.
b.
c.
npq 100(.5)(.5) = 5
(48 .5) 50
P(x ≤ 48) = P z
5
= P(z ≤ −.30)
= .5 − .1179 = .3821
P(50 ≤ x ≤ 65)
(65 .5) 50
(50 .5) 50
= P
z
5
5
= P(−.10 ≤ z ≤ 3.10)
= .0398 + .5000 = .5398
(70 .5) 50
P(x ≥ 70) = P z
5
= P(z ≥ 3.90)
= .5 − .5 = 0
d.
P(55 ≤ x ≤ 58)
(58 .5) 50
(55 .5) 50
= P
z
5
5
= P(.90 ≤ z ≤ 1.70)
= P(0 ≤ z ≤ 1.70) − P(0 ≤ z ≤ .90)
= .4554 − .3159 = .1395
e.
P(x = 62)
(62 .5) 50
(62 .5) 50
= P
z
5
5
= P(2.30 ≤ z ≤ 2.50)
= P(0 ≤ z ≤ 2.50) − (0 ≤ z ≤ 2.30)
= .4938 − .4893 = .0045
f.
P(x ≤ 49 or x ≥ 72)
(49 .5) 50
Pz
5
(72 .5) 50
Pz
5
= P(z ≤ −.10) + P(z ≥ 4.30)
= (.5 − .0398) + (.5 − .5) = .4602
4.197
x is normal random variable with μ = 40, σ2 = 36, and σ = 6.
a.
P(x ≥ x0) = .10
So, A = .5000 − .1000 = .4000.
(See part a.)
z0 = 1.28
To find x0, substitute the values into the z-score formula:
x μ
x 40
z0 = 0
1.28 = 0
x0 = 1.28(6) + 40 = 47.68
6
σ
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
251
252
Chapter 4
b.
P(μ ≤ x ≤ x0) = .40
Look up the area .4000 in the body of Table IV,
Appendix B; (take the closest value) z0 = 1.28.
To find x0, substitute the values into the z-score formula:
z0 =
c.
x0 μ
σ
1.28 =
x0 40
x0 = 40 + 6(1.28) = 47.68
6
P(x < x0) = .05
So, A = .5000 − .0500 = .4500.
Look up the area .4500 in the body of Table IV, Appendix B; z0 = −1.645. (.45 is halfway between
.4495 and .4505; therefore, we average the z-scores
1.64 1.65
= 1.645
2
z0 is negative since the graph shows z0 is on the left side of 0.
To find x0, substitute the values into the z-score formula:
z0 =
d.
x0 μ
σ
−1.645 =
x0 40
x0 = −1.645(6) + 40 = 30.13
6
P(x > x0) = .40
So, A = .5000 − .4000 = .1000.
Look up the area .1000 in the body of Table IV, Appendix
B; (take the closest value) z0 = .25.
To find x0, substitute the values into the z-score formula:
z0 =
e.
x0 μ
σ
.25 =
x0 40
x0 = 40 + 6(.25) = 41.5
6
P(x0 ≤ x < μ) = .45
Look up the area .4500 in the body of Table IV, Appendix
B; z0 = −1.645. (.45 is halfway between .4495 and .4505;
therefore, we average the z-scores
1.64 1.65
= 1.645
2
z0 is negative since the graph shows z0 is on the left side of 0.
4.198
a.
To find x0, substitute the values into the z-score formula:
x μ
x 40
z0 = 0
−1.645 = 0
x0 = 40 − 6(1.645) = 30.13
6
σ
First we must compute μ and σ. The probability distribution for x is:
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
x
1
2
3
4
μ = E(x) =
σ2 = E
253
p(x)
.3
.2
.2
.3
xp( x) = 1(.3) + 2(.2) + 3(.2) + 4(.3) = 2.5
(x μ)
2
=
(x μ )
2
p( x)
= (1 − 2.5) (.3) + (2 − 2.5)2(.2) + (3 − 2.5)2(.2)+ (4 − 2.52)(.3)
= 1.45
σ
1.45
μ x = μ = 2.5, σ x =
= .1904
n
40
b.
4.199
2
By the Central Limit Theorem, the distribution of is approximately normal. The sample size, n = 40,
is sufficiently large. Our answer does depend on n. If n is not sufficiently large, the Central Limit
Theorem would not apply.
By the Central Limit Theorem, the sampling distribution of is approximately normal.
μ x = μ = 19.6, σ x
a.
b.
c.
d.
3.2
68
= .388
19.6 19.6
P( x ≤ 19.6) = P z
= P(z ≤ 0) = .5
.388
(using Table IV, Appendix B)
19 19.6
P( x ≤ 19) = P z
= P(z ≤ −1.55) = .5 − .4394 = .0606
.388
(using Table IV, Appendix B)
20.1 19.6
P( x ≥ 20.1) = P z
= P(z ≥ 1.29) = .5 − .4015 = .0985
.388
(using Table IV, Appendix B)
20.6 19.6
19.2 19.6
P(19.2 ≤ x ≤ 20.6) = P
z
.388
.388
= P(−1.03 ≤ z ≤ 2.58) = .3485 + .4951 = .8436
(using Table IV, Appendix B)
4.200
Solution will vary. See page 1037 for Guided Solutions.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
254
4.201
Chapter 4
Given: μ = 100 and σ = 10
n
σ
n
1
5
10
20
30
40
50
10 4.472 3.162 2.236 1.826 1.581 1.414
The graph of σ / n against n is given here:
4.202
a.
In order for x to be a binomial random variable, the n trials must be identical. We can assume that
the process of selecting of a worker is identical from trial to trial. There are two possible outcomes a worker missed work due to a back injury or not. The probability of success must be the same from
trial to trial. We can assume that the probability of missing work due to a back injury is constant.
The trials must be independent of each other. We can assume that the outcome of one trials will not
affect the outcome of any other. Thus, x is a binomial random variable.
b.
From the information given in the problem, the estimate of p is .40.
c.
The mean is μ = E(x) = np = 10(.40) = 4.
The standard deviation is σ =
d.
np(1 p) 10(.40)(.60) 2.4 1 = 1.549
Using Table II, Appendix B, with n = 10 and p = .40,
P(x = 1) = P(x ≤ 1) − P(x ≤ 0) = .046 − .006 = .040
P(x > 1) = 1 − P(x ≤ 1) = 1 − .046 = .954
6
4.203
a.
p( x ) p(0) p(1) p(2) p(3) p(4) p(5)
i
i 1
.0102 .0768 .2304 .3456 .2592 .0778 1.0000
b.
P(x = 4) = .2592
c.
P(x < 2) = P(x = 0) + P(x = 1) = .0102 + .0768 = .0870
d.
P(x ≥ 3) = P(x = 3) + P(x = 4) + P(x = 5) = .3456 + .2592 + .0778 = .6826
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
255
6
e.
x p( x ) 0(.0102) 1(.0768) 2(.2304) 3(.3456) 4(.2592) 5(.0778)
μ E ( x)
i
i
i 1
0 .0768 .4608 1.0368 1.0368 .3890 3.0002
On the average, 3 out of every 5 dentists will use nitrous oxide.
4.204
a.
Using Table IV, Appendix B, with μ = 8.72 and σ = 1.10,
6 8.72
P(x < 6) = P z
= P(z < −2.47) = .5 − .4932 = .0068
1.10
Thus, approximately .68% of the games would result in fewer than 6 hits.
4.205
b.
The probability of observing fewer than 6 hits in a game is p = .0068. The probability of observing 0
hits would be even smaller. Thus, it would be extremely unusual to observe a no hitter.
a.
For this problem, c = 0 and d = 1.
1
1
(0 x 1)
f(x) = d c 1 0
0
otherwise
cd
0 1
=
= .5
2
2
(d c) 2
(1 0) 2
1
=
=
= .0833
σ2 =
12
12
12
σ = .0833 = .289
μ=
4.206
b.
P(.2 < x < .4) = (.4 − .2)(1) = .2
c.
P(x > .995) = (1 − .995)(1) = .005. Since the probability of observing a trajectory greater than .995 is
so small, we would not expect to see a trajectory exceeding .995.
a.
In the problem, it is stated that E(x) = .03. This is also the value of λ.
σ2 = λ = .03
b.
c.
The experiment consists of counting the number of deaths or missing persons in a three- year
interval. We must assume that the probability of a death or missing person in a three-year period is
the same for any three-year period. We must also assume that the number of deaths or missing
persons in any three-year period is independent of the number of deaths or missing persons in any
other three-year period.
λ 1e -λ .031e -.03
P(x = 1) =
= .0291
1!
1!
P(x = 0) =
λ 0e - λ
0!
.030e -.03
= .9704
0!
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256
4.207
Chapter 4
a.
μ x = μ = 89.34; σ x =
σ
n
=
7.74
35
= 1.31
b.
c.
d.
4.208
a.
88 89.34
P( x > 88) = P z
= P(z > −1.02) = .5 + .3461 = .8461
1.31
(using Table IV, Appendix B)
87 89.34
P( x < 87) = P z
= P(z < −1.79) = .5 − .4633 = .0367
1.31
(using Table IV, Appendix B)
Let x = number of trees infected with the Dutch elm disease in the two trees purchased. For this
problem, x is a hypergeometric random variable with N = 10, n = 2, and r = 3.
The probability that both trees will be healthy is:
r N r 3 10 3
3! 7!
x n x 0 2 0
0!3!
2!5! 1(21) = .467
P(x = 0) =
10!
45
N
10
n
2
2!8!
b.
The probability that at least one tree will be infected is:
P(x ≥ 1) = 1 − P(x = 0) = 1 − .467 = .533.
4.209
Let x = interarrival time between patients. Then x is an exponential random variable with a mean of 4
minutes.
a.
P(x < 1) = 1 − P(x ≥ 1)
= 1 − e−1/4
= 1 − e−.25
= 1 − .778801
= .221199
b.
Assuming that the interarrival times are independent,
P(next 4 interarrival times are all less than 1 minute)
= {P(x < 1)}4
= .2211994
= .002394
P(x > 10) = e−10/4
= e−2.5
= .082085
c.
4.210
a.
We must assume that the trials are identical, the probability of success is constant from trial to trial,
and the trials are independent of each other.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
b.
257
From the problem, we estimate p to be .20. Using Table II, Appendix B, with n = 25 and p = .20,
P(x ≤ 10) = .994
c.
E(x) = np = 25(.20) = 5
σ=
np(1 p) 25(.20)(.80) 4 = 2
d.
μ ± 2σ 5 ± 2(2) 5 ± 4 (1, 9)
e.
Using Table II, Appendix B, with n = 25 and p = .20,
P(1 < x < 9) = P(x ≤ 8) − P(x ≤ 1) = .953 − .027 = .926
4.211
a.
Let x1 = repair time for machine 1. Then x1 has an exponential distribution with μ1 = 1 hour.
P(x1 > 1) = e−1/1 = e−1 = .367879 (using Table V, Appendix B)
b.
Let x2 = repair time for machine 2. Then x2 has an exponential distribution with μ2 = 2 hours.
P(x2 > 1) = e−1/2 = e−.5 = .606531 (using Table V, Appendix B)
c.
Let x3 = repair time for machine 3. Then x3 has an exponential distribution with μ3 = .5 hours.
P(x3 > 1) = e−1/.5 = e−2 = .135335 (using Table V, Appendix B)
Since the mean repair time for machine 4 is the same as for machine 3, P(x4 > 1) = P(x3 > 1) =
.135335.
d.
The only way that the repair time for the entire system will not exceed 1 hour is if all four machines
are repaired in less than 1 hour. Thus, the probability that the repair time for the entire system
exceeds 1 hour is:
P(Repair time entire system exceeds 1 hour)
= 1 − P((x1 ≤ 1) ∩ (x2 ≤ 1) ∩ (x3 ≤ 1) ∩ (x4 ≤ 1))
= 1 − P(x1 ≤ 1)P(x2 ≤ 1)P(x3 ≤ 1)P(x4 ≤ 1)
= 1 − (1 − .367879)(1 − .606531)(1 − .135335)(1 − .135335)
= 1 − (.632121)(.393469)(.864665)(.864665) = 1 − .185954 = .814046
4.212
Let x = demand for white bread. Then x is a normal random variable with μ = 7200 and σ = 300:
a.
P(x ≤ x0) = .94. Find x0.
x 7200
P(x ≤ x0) = P z 0
300
= P(z ≤ z0) = .94
A1 = .94 − .50 = .4400
Using Table IV and area .4400, z0 = 1.555.
x 0 7200
x 7200
1.555 0
x0 = 7666.5 ≈ 7667
300
300
If the company produces 7,667 loaves, the company will be left with more than 500 loaves if the
demand is less than 7,667 - 500 = 7167.
7167 7200
P(x < 7167) = P z
= P(z < −.11)
300
= .5 − .0438 = .4562 (from Table IV, Appendix B)
z0 =
b.
Thus, on 45.62% of the days the company will be left with more than 500 loaves.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
258
4.213
Chapter 4
a.
In order for the number of deaths to follow a Poisson distribution, we must assume that the
probability of a death is the same for any week. We must also assume that the number of deaths in
any week is independent of any other week.
The first assumption may not be valid. The probability of a death may not be the same for every
week. The number of passengers varies from week to week, so the probability of a death may
change. Also, things such as weather, which varies from week to week may increase or decrease the
chance of derailment.
b.
E(x) = λ = 20
σ = λ = 20 = 4.47
c.
The z-score corresponding to x = 4 is:
4 20
z=
= −3.58
4.47
Since this z-score is more than 3 standard deviations from the mean, it would be very unlikely that
only 4 or fewer deaths occur next week.
d.
Using Table III, Appendix B with λ = 20,
P(x ≤ 4) = 0.000
This probability is consistent with the answer in part c. The probability of 4 or fewer deaths is
essentially zero, which is very unlikely.
4.214
Let x = number of inches a gouge is from one end of the spindle. Then x has a uniform distribution with
f(x) as follows:
1
1
1
f ( x) d c 18 0 18
0
0 x 18
otherwise
In order to get at least 14 consecutive inches without a gouge, the gouge must be within 4 inches of either
end. Thus, we must find:
P(x < 4) + P(x > 14) = (4 − 0)(1/18) + (18 − 14)(1/18)
= 4/18 + 4/18 = 8/18 = .4444
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.215
259
Using MINITAB, the stem-and-leaf display is:
Stem-and-leaf of Time
Leaf Unit = 0.10
(26)
23
15
9
4
2
2
1
1
1
1
2
3
4
5
6
7
8
9
10
N
= 49
00001122222344444445555679
11446799
002899
11125
24
8
1
The data are skewed to the right, and do not appear to be normally distributed.
Using MINITAB, the descriptive statistics are:
Variable
Time
Variable
Time
N
49
Mean
2.549
Minimum
1.000
Median
1.700
Maximum
10.100
TrMean
2.333
Q1
1.350
StDev
1.828
SE Mean
0.261
Q3
3.500
x ± s 2.549 ± 1.828 (0.721, 4.377)
x ± 2s 2.549 ± 2(1.828) 2.549 ± 3.656 (±1.107, 6.205)
x ± 3s 2.549 ± 3(1.828) 2.549 ± 5.484 (−2.935, 8.033)
Of the 49 measurements, 44 are in the interval (0.721, 4.377). The proportion is 44/49 = .898. This is
much larger than the proportion (.68) stated by the Empirical Rule.
Of the 49 measurements, 47 are in the interval (−1.107, 6.205). The proportion is 47/49 = .959. This is close
to the proportion (.95) stated by the Empirical Rule.
Of the 49 measurements, 48 are in the interval (−2.935, 8.033). The proportion is 48/49 = .980. This is
smaller than the proportion (1.00) stated by the Empirical Rule.
This would imply that the data are not normal.
IQR = QU − QL = 3.500 − 1.350 = 2.15. IQR/s = 2.15/1.828 = 1.176. If the data are normally distributed,
this ratio should be close to 1.3. Since 1.176 is smaller than 1.3, this indicates that the data may not be
normal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
260
Chapter 4
Using MINITAB, the normal probability plot is:
Since this plot is not a straight line, the data are not normal.
All four checks indicate that the data are not normal.
4.216
a.
By the Central Limit Theorem, the sampling distribution of is approximately normal with
μ x = μ and σ x σ / n .
b.
Let μ = 18.5. Since we do not know σ we will estimate it with s = 6.
19.1 18.5
P( x 19.1) P z
= P(z ≥ 1.85) = .5 − .4678 = .0322
6 / 344
c.
Let μ = 19.5. Since we do not know σ we will estimate it with s = 6.
19.1 19.5
P( x 19.1) P z
= P(z ≥ −1.24) = .5 + .3925 = .8925
6 / 344
d.
If P( x ≥ 19.1) = .5, then the population mean must be equal to 19.1. (For a normal distribution, half
of the distribution is above the mean and half is below the mean.)
e.
If P( x ≥ 19.1) = .2, then the population mean is less than 19.1. We know the probability that is
greater than the mean is .5. Since P(x ≥ 19.1) = .2 which is less than .5, we know that 19.1 must be
to the right of the mean. Thus, the population mean must be less than 19.1.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.217
261
Let x equal the difference between the actual weight and recorded weight (the error of measurement). The
random variable x is normally distributed with μ = 592 and σ = 628.
a.
We want to find the probability that the weigh-in-motion equipment understates the actual weight of
the truck. This would be true if the error of measurement is positive.
0 592
P(x > 0) = P z
628
= P(z > −.94)
= .5000 + .3264
= .8264
b.
P(overstate the weight) = 1 − P(understate the weight)
= 1 − .8264
= .1736 (Refer to part a.)
For 100 measurements, approximately 100(.1736) = 17.36 or 17 times the weight would be
overstated.
c.
d.
400 592
P(x > 400) = P z
628
= P(z > −.31)
= .5000 + .1217
= .6217
We want P(understate the weight) = .5
To understate the weight, x > 0. Thus, we want to find μ so
that P(x > 0) = .5
0 μ
P(x > 0) = P z
= .5
628
μ
From Table IV, Appendix B, z0 = 0. To find μ, substitute into the z-score formula:
x μ
0μ
z0 = 0
0=
μ=0
628
σ
Thus, the mean error should be set at 0.
We want P(understate the weight) = .4
To understate the weight, x > 0. Thus, we want to find μ so that
P(x > 0) = .4
μ
A = .5 − .40 = .1. Look up the area .1000 in the body of Table IV, Appendix B, z0 = .25.
To find μ, substitute into the z-score formula:
z0 =
x0 μ
σ
.25 =
0μ
μ = 0 − (.25)628 = −157
628
Thus, the mean error should be set at −157.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
262
4.218
Chapter 4
a.
Let p1 = probability of an error = 1/100 = .01 and p2 = probability of an error resulting in a significant
problem = 1/500 = .002.
Let x = number of errors in 60,000 trials. Then E(x) = μ1 = np1 = 60,000(.01) = 600.
Let y = number of significant errors in 60,000 trials. Then E(y) = μ2 = np2 = 60,000(.002) = 120.
b.
σ = np2q2 = 60,000(.002)(.998) = 119.76
σ = 119.76 = 10.94
μ2 ± 3σ 120 ± 3(10.94) 120 ± 32.82 (87.18, 152.82)
Using Chebyshev's Rule, at least 88.9% of the observations will fall within 3 standard deviations of
the mean. We would expect the number of significant errors to fall between 87 and 153.
4.219
c.
We must assume that the trials are independent and that the probability of a significant error is
constant from trial to trial.
a.
μ = n ⋅ p = 25(.05) = 1.25
σ = npq 25(.05)(.95) = 1.09
Since μ is not an integer, x could not equal its mean.
b.
The event is (x ≥ 5). From Table II with n = 25 and p = .05:
P(x ≥ 5) = 1 − P(x ≤ 4) = 1 − .993 = .007
4.220
c.
Since the probability obtained in part b is so small, it is unlikely that 5% applies to this agency. The
percentage is probably greater than 5%.
a.
By the Central Limit Theorem, the sampling distribution of x is approximately normal since n > 30
and
σ
15
σx
μ x = μ = 840
= 2.1213
n
50
b.
830 840
P( x ≤ 830) = P z
= P(z ≤ −4.71) ≈ .5 − .5 = 0
2.1213
c.
Since the probability of observing a mean of 830 or less is extremely small (≈0) if the true mean is
840, we would tend to believe that the mean is not 840, but something less.
d.
By the Central Limit Theorem, the sampling distribution of is approximately normal since n > 30
and
σ
45
σx
μ x = μ = 840
= 6.3640
n
50
830 840
P( x ≤ 830) = P z
= P(z ≤ −1.57) ≈ .5 − .4418 = .0582
6.3640
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.221
263
Let x = number of grants awarded to the north side in 140 trials. The random variable x has a
hypergeometric distribution with N = 743, n = 140, and r = 601.
a.
μ = E(x) =
nr 140(601)
= 113.24
N
743
r ( N r ) n( N n)
σ2 =
N ( N 1)
2
601(743 601)140(743 140)
7432 (743 1)
= 17.5884
σ = 17.5884 = 4.194
b.
If the grants were awarded at random, we would expect approximately 113 to be awarded to the
north side. We observed 140. The z-score associated with 140 is:
z=
xμ
σ
140 113.24
= 6.38
4.194
Because this z-score is so large, it would be extremely unlikely to observe all 140 grants to the north
side if they are randomly selected. Thus, we would conclude that the grants were not randomly
selected.
4.222
Let x = length of time a bus is late. Then x is a uniform random variable with probability distribution:
1
(0 x 20)
f(x) = 20
0 otherwise
μ=
b.
1 1
P(x ≥ 19) = (20 − 19) ⋅
= .05
20 20
c.
4.223
0 20
= 10
2
a.
It would be doubtful that the director’s claim is true, since the probability of the being more than 19
minutes late is so small.
We know from the Empirical Rule that almost all the observations are larger than
μ − 2σ. (≈ 95% are between μ − 2σ and μ + 2σ). Thus μ − 2σ > 100.
For the binomial, μ = np = n(.4) and σ =
npq n(.4)(.6) =
.24n
μ − 2σ > 100 .4n − 2 .24n > 100 .4n − .98 n − 100 > 0
Solving for
n
n , we get:
.98 .98 2 4(.4)(100) .98 12.687
2(.4)
.8
n = 17.084 n = 17.0842 = 291.9 ≈ 292
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
264
4.224
Chapter 4
Let x = tensile strength of a particular metal part. Then x is a normal random variable with
mean = μ = 25 and standard deviation = σ = 2. The tolerance limits are 21 and 30.
21 25
P( x 21) P z
P( z 2) .5 .4772 .0228 (Using Table IV, Appendix B).
2
30 25
21 25
P(21 x 30) P
z
P (2 z 2.5) .4772 .4938 .9710
2
2
30 25
P( x 30) P z
P ( z 2.5) .5 .4938 .0062
2
E(Profit) = −$2(.0228) + $10(.9710) − $1(.0062) = −$.0456 + $9.71 − $.0062 = $9.66
4.225
Even though the number of flaws per piece of siding has a Poisson distribution, the Central Limit Theorem
implies that the distribution of the sample mean will be approximately normal with μ x = μ = 2.5 and
σx
σ
n
2.5
35
.2673 . Therefore,
2.1 2.5
P( x > 2.1) + P z
= P(z > −1.50) = .5 + .4332 = .9332 (using Table IV, Appendix B)
2.5 / 35
4.226
Let x = number of packets out of 4 that contain cocaine. Then x is an approximate binomial random
variable with n = 4 and p= 331/496 = .667.
4
P( x 4) .667 4 (.333)4 4 .6674 .197
4
Let y = number of packets out of 2 not containing cocaine, given the first 4 contained cocaine. Then y is
an approximate binomial random variable with n = 2 and p = 165/492 = .335.
2
P( y 2) .3352 (.665)2 2 .3352 .112
2
The probability that the first 4 packets contained cocaine and the next 2 did not is:
P( x 4) P ( y 2) .197(.112) .022
Thus, it is very unlikely that the first 4 packets contained cocaine and the last 2 did not.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Random Variables and Probability Distributions
4.227
265
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: INSOMNIA
Variable
INSOMNIA
N
40
Mean
5.935
StDev
5.392
Minimum
1.300
Q1
2.250
Median
3.650
Q3
7.675
Maximum
22.800
To see if the mean time to fall asleep for those taking melatonin is different from those taking a placebo,
we will see how unusual it is to observe a sample mean of 5.935 if the true mean is 15.
5.935 15
P( x 5.935) P z
P ( z 10.63) .5 .5 0 (Using Table IV, Appendix B.)
5.392
40
Since this probability is so small, we would infer that the mean time to fall asleep for those taking
melatonin is not 15 minutes, but something less than 15 minutes.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
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