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Chapter 06 - Continuous Random Variables
CHAPTER 6—Continuous Random Variables
6.1
Intervals of values.
6.2
By finding areas under the curve.
6.3
f(x)  0 for all x; area under the curve equals 1.
6.4
Relative likelihood that x will be near the given point.
6.5
When a variable has a rectangular distribution over a certain interval.
6.6
MTB > cdf cl;
SUBC> uniform 2.0 8.0.
CONTINUOUS UNIFORM ON 2.0 TO 8.0
K P( X LESS OR = K)
0.0000
0.0000
0.5000
0.0000
1.000 0
0.0000
1.5000
0.0000
2.0000
0.0000
2.5000
0.0833
3.0000
0.1667
3.5000
0.2500
4.0000
0.3333
4.5000
0.4167
5.0000
0.5000
5.5000
0.5833
6.0000
0.6667
6.5000
0.7500
7.0000
0.8333
7.5000
0.9167
8.0000
1.0000
a.
f ( x) 
1
1
 for 2  x  8
8–2 6
= 0 otherwise
b.
Graph not included in this manual.
c.
P(3  x  5) = P(x  5) – P(x  3) = .5 – .1667 = .3333
d.
P(1.5  x  6.5) = P(x  6.5) – P(x  1.5) = .75 – 0 = .75
e.
cd 28

5
2
2
(d – c) 2 (8 – 2) 2
 x2 

3
12
12
x 
 x   x2  3  1.732
6-1
Chapter 06 - Continuous Random Variables
f.
6.7
6.8
[  x  2 x ]  [5  2(1.732 )]  [1.536,8.464 ]
P(1.536  x  8.464) = P(x  8) = 1
f ( x) 
a.
1
1
1


d  c 175  50 125
f ( x) 
1
1
 for 0  x  6.
60 6
= 0 otherwise
b.
f(x)
1/6
0
6.9
6 min
c.
1
P(2  x  4)  (4 – 2)  .3333
6
d.
1
P(3  x  6)  (6 – 3)  .5
6
e.
1 1
P{0  x  2} or {5  x  6}  2   1  .5
6 6
a.
b.
cd 06

3
2
2
(d – c) 2 (6 – 0) 2
 x2 

3
12
12
 x  3  1.732
x 
[  x   x ]  [3  1.732 ]  [1.268, 4.732 ]
1
P(1.268  x  4.732)  (4.732 – 1.268)  .5773
6
6.10
6.11
1
f ( x)  d  c
a.
1
2
1
 50
1
2

2
5
for 120  x  140
6-2
Chapter 06 - Continuous Random Variables
f ( x) 
b.
1
1
1


d  c 140  120 20
f(x)
1/20
120
6.12
140 min
c.
P(125  x  135) = 10 (1/20) = .5
d.
P(x  135) = 5 (1/20) = .25
a.
b.
c  d 120  140

 130
2
2
d  c 140  120
x 

 5.7735
12
12
x 
 x   x  130  5.7735  124 .2265
 x   x  130  5.7735  135 .7735
P(124 .2265  x  135 .7735 )  (135 .7735  124 .2265 ) x
6.13
6.14
1
f ( x)  d  c
a.
b.
f ( x) 
1
2
1
 17  5
1
2

2 1

12 6
1
1
 for 3  x  6.
6–3 3
f(x)
1/3
3
c.
6 inches
P(x  4) = 1 – P(x < 4) = 1 – .3333 = .6667
P(x  5) = 1 – P(x < 5) = 1 – .6667 = .3333
1
P( x  4.5)  (4.5 – 3)  .5
 3
6-3
1
 .57735
20
Chapter 06 - Continuous Random Variables
6.15
6.16
a.
x 
b.
x 
c  d 36

 4.5
2
2
d –c

6–3
.8660
12
12
[  x  2 x ]  [4.5  2(.8660 )]  [2.768, 6.232 ]
P(2.768  x  6.232) = P(3  x  6) = 1
[x  x ]  [4.5.8660]  [3.634, 5.366]
1
P(3.634  x  5.366)  (5.366 – 3.634)  .5773
 3









Entire family of normal distributions
Highest point on curve is mean, median and mode
Distribution is symmetrical
Tails go to infinity in both directions



Area right of  = Area left of  = 0.5
a.
center
b.
spread
6.17

6.18
68.26%, 95.44%, 99.73%
6.19
Subtract the mean and divide the result by the standard deviation; tells us the number of
standard deviations the value is above or below the mean.
6.20
a.
x equals the mean
b.
x greater than the mean
c.
x less than the mean
6.21
The normal table provides the areas under the standard normal curve (the distribution of the z
values).
6-4
Chapter 06 - Continuous Random Variables
6.22
a.
 =3
 =6
20
X
b.
20
30
c.
100
6.23
200
a.
z
25 – 30
 –1 ; x is one standard deviation below the mean.
5
b.
z
15 – 30
 –3 ; x is three standard deviations below the mean.
5
c.
z
30 – 30
 0 ; x is equal to the mean.
5
6-5
Chapter 06 - Continuous Random Variables
6.24
d.
z
40 – 30
 2 ; x is two standard deviations above the mean.
5
e.
z
50 – 30
 4 ; x is four standard deviations above the mean.
5
a.
P(0  z  1.5) = 0.9332 – 0.5000 = .4332
b.
P(z  2) = 1 – .9772 = .0228
6-6
Chapter 06 - Continuous Random Variables
c. P(z  1.5) =.9332
d.
P(z  –1) = 1 – .1587 =.8413
e.
P(z  –3) = .00135
f.
P(–1  z  1) = .8413 – .1587 = .6826
6-7
Chapter 06 - Continuous Random Variables
6.25
6.26
g.
P(–2.5  z  .5) = .6915 – .0062 = .6853
h.
P(1.5  z  2) = .9772 – .9332 = .0440
i.
P P(–2  z  –.5) = .3085 – .0228 = .2857
a.
z.01  2.33
b.
z .05  1.645
c.
z.02  2.054 or 2.05 with rounding
d.
– z .01  –2.33
e.
– z .05  –1.645
f.
– z .10  –1.28
Restate each probability in terms of the standard normal random variable z.
x –  x – 1000
z


100
Then use the table to solve.
6-8
Chapter 06 - Continuous Random Variables
a.
P(1000  x  1200) = P(0  z  2) = .9772 – .5 = .4772
b.
P(x > 1257) = P(z > 2.57) = 1 – .9949 = .005
c.
P(x < 1035) = P(z < .35) = .6368
d.
P(857  x  1183) = P(–1.43  z  1.83) = .9664 – .0764 = .8900
e.
P(x  700) = P(z  – 3) = .00135
f.
P(812  x  913) = P(–1.88  z  –.87) = .1922 – .0301 = .1621
g.
P(x > 891) = P(z > –1.09) = 1 – .1379 = .8621
h.
P(1050  x  1250) = P(.5  z  2.5) = .9938 – .6915 = .3023
First find the z-value from the table that makes the statement true. Then calculate x using
the formula:
x = z +  = z(100) + 500
6.27
6.28
a.
P(x  696) = .025
b.
P(x  664.5) = .05
c.
P(x < 304) = .025
d.
P(x  283) = .015
e.
P(x < 717) = .985
f.
P(x > 335.5) = .95
g.
P(x  696) = .975
h.
P(x  700) = .0228
i.
P(x > 300) = .9772
a.
b.
z
x–


x – 100
16
6-9
Chapter 06 - Continuous Random Variables
6.29
c.
(1) P(x > 140) = P(z > 2.5) = 1 – .9938 = .0062
(2) P(x < 88) = P(z < –.75) = .2266
(3) P(72 < x < 128) = P(–1.75 < z < 1.75) = .9599 – .0401= .9198
(4) P(–1.5  z  1.5) = .9332 – .0668 = .8664
d.
P(x > 136) = P(z > 2.25) = 1 – .9878 = .0122; 1.22%
a.
(1) P(x  959) = P(z  2.12) = .9830
(2) P(x > 1004) = P(z > 2.72) = 1 – .9967 = .0033
(3) P(x < 650) + P(x > 950) = P(z < –2) + P(z > 2) = .0228 + (1 – .9772) = .0456
b.
P(x > 947) = P(z > 1.96) = .025
order  100
 1.96
16
order = 947 boxes of cereal
6.30
6.31
6.32
Restate each probability in terms of z. Then use the table to solve.
a.
b.
c.
d.
e.
f.
g.
h.
P(7  x  9) = P(–2.0 ≤ z ≤ 2.0) = .9772 – .0228 = .9544
P(8.5  x  9.5) = P(1 ≤ z ≤ 3) = .9987 – .8413 = .1574
P(6.5  x  7.5) = P(–3 ≤ z ≤ –1) = .1587 – .00135 = .15735
P(x  8) = P(z  0) = 1 – .5 = .5
P(x  7) = P(z  –2) = .0228
P(x  7) = P(z  –2) = 1 – .0228 = .9772
P(x  10) = P(z  4) = 1.0 (approximately)
P(x > 10) = P(z > 4) = 0 (approximately)
a.
P(x  27) = P(z  –3.00) = .5 – .00135 = .49865
b.
Claim is probably not true, because the probability is very low of randomly purchasing a
car with 27 mpg if the mean is actually 30 mpg.
Common stocks:  = 12.4,  = 20.6; let x = stock.
Tax free municipal bond:  = 5.2,  = 8.6; let y = bond.
a.
b.
c.
d.
e.
f.
P(x > 0) = P(z > –.60) = 1 – .2743 = .7257
P(y > 0) = P(z > –.60) = 1 – .2743 = .7257
P(x > 10) = P(z > –.12) = 1 – .4522 = .5478
P(y > 10) = P(z > .56) = 1 – .7123 = .2877
P(x  –10) = P(z  –1.09) = .1379
P(y  –10) = P(z  –1.77) = .0384
6-10
Chapter 06 - Continuous Random Variables
6.33
6.34
15.95 – 16.0024 
16.05 – 16.0024 


P( x  15.95)  P( x  16.05)  P z 
  P z 

.02454
.02454




= P(z < –2.14) + P(z > 1.94) = .0162 + (1 – .9738) = .0424
P(x  k) = .02
k–
z

k – 40,000
4000
k = 31,784
Approximately 31,784 miles
– 2.054 
Note: if use rounded z value of –2.05, then k =31,800
6.35
a.
10%, 90%
k 
z

 1.28 
k  12.4
20.6
k  13.968
b.
Q1 :  0.67 
k  12.4
20.6
k  1.402
Q3 : 0.67 
k  12.4
20.6
k  26.202
6.36
z
x–

, –1
63 – 

, 1.5 
93 – 

Solve the two equations for the two unknowns:  = 75,  = 12
6.37
6.38
a.
[ ± 2.33]
b.
[50.575 ± 2.33(1.6438)] = [46.745, 54.405]
a.
 + 3 = 5.2 + 3(8.6) = 31%
b.
31  12.4 

P( x  31)  P z 
  Pz .90  1  .8159 .1841
20.6 

6-11
Chapter 06 - Continuous Random Variables
6.39
2500  0 

  P( z .5) 1 .6915 .3085
5000 

2500  0 

Process B: P( x  2500)  P z  10,000   P( z .25) 1 .5987 .4013


Process A: P( x  2500)  P z 
a.
Process B is investigated more often.
2500  7500 

  P( z  1) 1  .1587 .8413
5000


2500  7500 

Process B: P( x  2500)  P z  10,000   P( z   .5) 1 .3085 .6915


Process A: P( x  2500)  P z 
b.
Process A is investigated more often.
c.
Process B will be investigated more often.
d.
P(x > k) = .3085 implies that z 
k 0
 .5 . Thus k = 5000.
10,000
Investigate if cost variance exceeds $5000.
5000  7500 

P( x  5000)  P z 
  P( z  .25)  1  .4013  .5987
10,000 

6.40
P(x  k) = .33
k
z

k  3000
500
k = 2780
$2780
 .44 
6.41
656  

 .44
896  

 1.96
656 –  = –.44
896 –  = 1.96
– = –.44 – 656
– = 1.96 – 896
 = .44 + 656
 = –1.96 + 896
.44 + 656 = –1.96 + 896
2.4 = 896 – 656
2.4 = 240
 = 100
 = .44 + 646
 = .44(100) + 656 = 44 + 656 = 700
6.42
Binomial tables are often unavailable for large values of n.
6.43
Both np and n(1 – p) exceed 5.
6-12
Chapter 06 - Continuous Random Variables
6.44
Create an interval that includes the integer (or integer interval) to be able to find the area
under the curve. This is necessary because a discrete distribution is being approximated by a
continuous distribution.
6.45
a.
np = (200)(.4) = 80
n(1 – p) = (200)(.6) = 120
both  5
6.46
b.
  np  (200 )(.4)  80,  npq  48  6.9282
Rounding z to 2 decimal places
(1) P(x = 80) = P(79.5  x  80.5) = P(–.072  x  .072) = .0576
.0558
(2) P(x  95) = P(x  95.5) = P(z  2.237) = .9874
.9875
(3) P(x < 65) = P(x  64.5) = P(z  –2.237) = .0126
.0125
(4) P(x  100) = P(x  99.5) = P(z  2.8146) = .0024
.0025
(5) P(x > 100) = P(x  100.5) = P(z  2.959) = .0015
a.
Both np and n(1 – p) exceed 5.
b.
 = 100,  = 7.0711
See the methods outlined in the solution to Exercise 5.45.
P(x = 80) = .001
P(x  95) = .2623
P(x < 65) = .0000
P(x  100) = .5282
P(x > 100) = .4718
6.47
6.48
a.
(1) Both np and n(1 – p) exceed 5.
(2)   np  (1000 )(.2)  200,  npq  160  12.6491
(3) P(x  150) = P(x  150.5) = P(z  –3.913) = less than .001
b.
No. If the claim were true, the probability of observing this survey result is less than
.001.
a.
  (205)(.65)  133.25,  46.6375  6.8292
117.5  133.25 

P( x  117 )  P z 
  .0105 , rounding z to 2 decimal places gives .0104
6.8292


6.49
b.
No
a.
  (250)(.05)  12.5,  11.875  3.446
39.5  12.5 

P( x  40)  P z 
  0 (approximately)
3.446 

b.
No
6-13
Chapter 06 - Continuous Random Variables
6.50
  (2000 )(.20)  400,  320  17.8885
P(x  st) = .99
st  
z

st  400
17.8885
st = 441.7
442 units
2.33 
6.51
Explanations will vary.
6.52
f ( x)  e x for x  0 ,
= 0 otherwise
6.53
Explanations will vary.
6.54
a.
f ( x)  2e 2 x for x  0.
b.
Graph not included in this manual.
c.
P( x  1)  P(0  x  1)  e 0  e 2  1 .1353 .8647
d.
P(.25  x  1)  e 2(.25)  e 2(1) .4712
e.
P( x  2)  e 2( 2) .0183
f.
x 
g.
 x  2 x   [.5  2(.5)]  [.5,1.5]
1 1
1
1 1
  .5, 2x  2  .25, x    .5
 2
 2

P(.5  x  1.5)  e 2(0)  e 2(1.5)  .9502
6.55
a.
f ( x)  3e 3 x for x  0.
b.
Graph not included in this manual.
c.
P(x  1) = .9502
d.
P(.25  x  1) = .4226
e.
P(x  2) = .0025
f.
 x  , x2  , x 
1
3
1
9
1
3
6-14
Chapter 06 - Continuous Random Variables
6.56
g.
 x  2 x    1 ,1, P  1  x  1  .9502
a.
f ( x) 
b.
c.
6.57

7 7 x / 15
e
for x  0.
15
Graph not included in this manual.
P(a  x  b)  e a  e b  e 7 a / 15  e 7b / 15
(1) P(1  x  2) = .2338
(2) P(x < 1) = .3729
(3) P(x > 3) = .2466
(4) P(.5  x  3.5) = .5966
1
2
15 2  1 
225
1 15
, x    
, x  
7

49

7
 
d.
x 
e.
 x   x   15  15   [0, 4.2857 ], P(0  x  4.2857 )  .8647
a.
b.
c.
6.58
 3   3


7
7
 x  2 x   15  2 15   [2.1429 , 6.4286 ], P(0  x  6.4286 ) .9502
 7 
7
2
f ( x)  e 2 x / 3 for x  0.
3
Graph not included in this manual.
P(a  x  b)  e a  e b  e 2a / 3  e 2b / 3
(1) P(x  3) = .8647
(2) P(1  x  2) = .2498
(3) P(x > 4) = .0695
(4) P(x < .5) = .2835
1
2
a.
 x  (500)  250,  
b.
f ( x) 
1
x

1
250
1  x / 250
e
for x  0.
250
c.
Graph not included in this manual.
d.
P( x  5)  1  e 5 / 250  .0198
e.
P(100  x  300)  e 100/ 250  e 300/ 250  .3691
6-15
Chapter 06 - Continuous Random Variables
6.59
f.
Probably not; the probability of this result is quite small (.0198) if the claim is true.
a.
 1
2
(1) P( x  2)  e  .1353
1
2
(2) P(1  x  2)  e  e  .2325
.25
(3) P( x .25)  1  e  .2212
b.
Probably not; the probability of this happening is .2212 (which is not terribly small).
6.60
See the steps on Pg 6-34?????? titled Normal Probability Plots
6.61
That the data approximately follows a normal distribution
a & b.
7.524
11.070
18.211
26.817
36.551
41.286
49.312
57.283
72.814
90.416
135.540
190.250
c.
i/(n+1)
0.0769
0.1538
0.2308
0.3077
0.3846
0.4615
0.5385
0.6154
0.6923
0.7692
0.8462
0.9231
z-value
-1.43
-1.02
-0.74
-0.50
-0.29
-0.10
0.10
0.29
0.50
0.74
1.02
1.43
Data is skewed because plot does not show up as a straight line.
Normal Prob. Plot
200.000
150.000
100.000
50.000
0.000
-1
.4
3
-1
.0
2
-0
.7
4
-0
.5
0
-0
.2
9
-0
.1
0
0.
10
0.
29
0.
50
0.
74
1.
02
1.
43
Income
(1000's)
Income
6.62
z-score
6-16
Chapter 06 - Continuous Random Variables
6.63
Plot below shows data does not follow a normal distribution since it is not a straight line.
Normal Curve Plot
12
10
Ratings
8
6
4
2
0
-3.0
-2.0
-1.0
0.0
1.0
2.0
3.0
Normal Score
6.64
Data is approximately normal because plot is close to a straight line.
Normal Curve Plot
33.5
33.0
32.5
MPG
32.0
31.5
31.0
30.5
30.0
29.5
-3.0
-2.0
-1.0
0.0
1.0
2.0
3.0
Normal Score
15.95  16
)  P( z  2.5)  .0062; .62%
.02
6.65
P( x  15.95)  P( z 
6.66
a.
1  P(71  x  76)  1  P(
b.
P(x  74) = P(z  .5) = 1 – .6915 = .3085
c.
P(x < 70.5) = P(z < –3) = .00135
71  73.5
76  73.5
z
)  1  P(2.5  z  2.5)
1
1
 1  .9876  .0124
6-17
Chapter 06 - Continuous Random Variables
6.67
a.
P( x  80)  P( z 
b.
P( x  k )  .05
k
z
80  100
)  P( z  1.25)  1  .1056  .8944
16

k  100
16
Minimum score  73.68
 1.645 
6.68
a.
f ( x) 
b.
1
1

1
.5  (.5) .5  .5
for  .5  x  .5
f(x)
1
–.5
.5 cents
c.
P({x > .3} or {x < –.3}) = .2 + .2 = .4
d.
P({x > .1} or {x < =.1}) = .4 + .4 = .8
e.
c  d  .5  .5

0
2
2
(d  c) 2
(.5  (.5)) 2
x 

 .2887
12
12
f.
[  x   x ]  [0  .2887 ]  [.2887 ,. 2887 ]
P(.2887  x  .2887 )  (.2887  (.2887 ))(1)  .5774
a.
P(x  3.5) = P(z  –1.25) = 1-.1056 = .8944
b.
P(x  6) = P(z  0.83) = .7967
c.
P(3.5  x  6) = .7967 – .1056 = .6911
a.
10%, 90%, approximately 3.462
ux 
6.69
6.70
P( x  k )  .10
k
z

 1.282 
k 5
, k  3.4616
1.2
Note: if use z = –1.28, k = 3.464
6-18
Chapter 06 - Continuous Random Variables
b.
Follow the methods outlined in part a, or compute the inverse cdf in MINITAB.
Q1 = 4.196, Q3 = 5.804
6.71
P( x  k )  .025
k
z

1.96 
6.72
Set lowest passing score to 298
k  200
, k  298
50
P(x < 15) = .004
15  
z

15  
.02
 = 15.053
Set to 15.053 inches.
 2.65 
6.73
6.74
300.5  292.5 

P( x  300.5)  P z 
  P( z  1.48)  .9306
5.40833 


1
x

4
1

1000 250
a.
P( x  400)  e 400 / 250 .2019
b.
P(0  x  100)  e 0  e 100 / 250  1 .6703 .3297
6-19
Chapter 06 - Continuous Random Variables
6.75
a, b. The probabilities below were computer calculated. Answers obtained using the normal
table may be slightly different.
Probability
of a return
Fixed annuities
Cash
equivalents
U.S. treasury
bonds
U.S. corporate
bonds
<0
.0000
.0000
.0708
.1190
> 5%
1.0000
.9996
.7374
.7077
> 10%
.0009
.0025
.4205
.4663
> 20%
.0000
.0000
.0305
.0890
> 50%
.0000
.0000
.0000
.0000
Non-U.S.
government bonds
Domestic large
cap stocks
International
equities
Domestic MidCap
stocks
<0
.1469
.2206
.2061
.2266
> 5%
.7151
.6695
.7004
.6826
> 10%
.5361
.5445
.5926
.5793
>20%
.1937
.2940
.3637
.3633
> 50%
.0001
.0062
.0180
.0228
Domestic
small cap stocks
c.
<0
.2483
> 5%
.6755
> 10%
.5894
> 20%
.4081
> 50%
.0540
The probability of a return greater than 50% is:
(1) essentially zero for fixed annuities, cash equivalents, U.S., treasury bonds, U.S.
investment grade corporate bonds, non-U.S. government bonds, and domestic large
cap issues.
(2) greater than 1% for international equities, domestic MidCap stocks, and domestic
small cap stocks.
(3) greater than 5% for domestic small cap stocks.
d.
The probability of a loss is:
(1) essentially zero for fixed annuities and cash equivalents.
6-20
Chapter 06 - Continuous Random Variables
(2)
(3) greater than 1% for all investment-classes except fixed annuities and cash
equivalents.
(4) greater than 10% for all investment classes except fixed annuities, cash equivalents,
and U.S. treasury bonds.
(5) greater than 20% for domestic large cap stocks, international equities, domestic
MidCap stocks, and domestic small cap stocks.
6.76
P(x  984,000)  P(z 
6.77
25  15 10 2
 
25  10 15 3
6.78
x = 60 sec,  = 1
6.79
984,000 - 800,000
 P(z  2.3)  1  .9893  .0107
80,000
a.
P( x  1.5 min)  e 1.5  .2231
b.
P( x  2 min)  e 2  .1353
 = 4.15,  = .5
a.
P( x  5.4)  P( z 
5.4  4.15
)  P( z  2.5)  1  .9983  .0062
.5
b.
P( x  4.4)  P( z 
4.4  4.15
)  P( z  .5)  .6915
.5
c.
P ( x  k )  .05
k
z

k  4.15
.5
k  3.3275
 1.645 
6-21
Chapter 06 - Continuous Random Variables
6.80
 = np = (400)(.50) = 200
  npq  (400 )(.5)(.5)  100  10
a.
P(x  180)  P( z 
b.
Yes
180.5  200
)  P( z  1.95)  1  .9744  .0256
10
6.81
P(20  x  30)  P(
20  26
30  26
z
)  P(1.5  z  1.0)  .8413  .0668  .7745
4
4
6-22