Download Chapter Five

Chapter Five
5.1
Random variable: A variable whose value is determined by the outcome of a random experiment is
called a random variable. An example of this is the income of a randomly selected family.
Discrete random variable: A random variable whose values are countable is called a discrete random
variable. An example of this is the number of cars in a parking lot at any particular time.
Continuous random variable: A random variable that can assume any value in one or more intervals
is called a continuous random variable. An example of this is the time taken by a person to travel by
car from New York City to Boston.
5.3
a. a discrete random variable
b.
a continuous random variable
c. a continuous random variable
d.
a discrete random variable
e. a discrete random variable
f.
a continuous random variable
5.5
The number of cars x that stop at the Texaco station is a discrete random variable because the values of
x are countable: 0, 1, 2, 3, 4, 5 and 6.
5.7
The two characteristics of the probability distribution of a discrete random variable x are:
1. The probability that x assumes any single value lies in the range 0 to 1, that is, 0  P( x)  1 .
2. The sum of the probabilities of all values of x for an experiment is equal to 1, that is:
5.9
a. P(x = 1) = .17
b. P(x  1) = P(0) + P(1) = .03 + .17 + = .20
c. P(x  3) = P(3) + P(4) + P(5) = .31 + .15 + .12 = .58
d. P(0  x  2) = P(0) + P(1) + P(2) = .03 + .17 + .22 = .42
e. P(x < 3) = P(0) + P(1) + P(2) = .03 + .17 + .22 = .42
f. P(x >3) = P(4) + P(5) = .15 + .12 = .27
g. P(2  x  4) = P(2) + P(3) + P(4) = .22 + .31 + .15 = .68
71
 P( x)  1 .
72
5.11
Chapter Five
a.
b. i. P(exactly 3) = P(3) = .25
ii. P(at least 4) = P(x  4) = P(4) + P(5) + P(6) = .14 + .07 + .03 = .24
iii. P(less than 3) = P(x< 3) = P(0) + P(1) + P(2) = .10 + .18 + .23 = .51
iv. P(2 to 5) = P(2) + P(3) + P(4) + P(5) = .23 + .25 + .14 + .07 = .69
a.
x
1
2
3
4
5
P(x)
8 / 80 = .10
20 / 80 = .25
24 / 80 = .30
16 / 80 = .20
12 / 80 = .15
30
P(x)
5.13
20
10
0
b. The probabilities listed in the table of part
a are approximate because they are
1
2
3
4
5
Number of Systems Installed
obtained from a sample of 80 days.
c. i. P(x = 3) = .30
ii. P(x  3) = P(3) + P(4) + P(5) = .30 + .20 + .15 = .65
iii. P(2  x  4) = P(2) + P(3) + P(4) = .25 + .30 + .20 = .75
iv. P(x ≤4) = P(2) + P(3) + P(4) = .10 + .25 + .30 = .65
5.15
Let Y = owns a cell phone and N = does not own a cell phone.
Then P(Y) = .64 and P(N) = 1 – .64 = .36
Let x be the number of adults in a sample of two who own a cell phone. The following table lists the
probability distribution of x. Note that x = 0 if neither adult owns a cell phone, x = 1 if one adult owns
a cell phone and the other does not, and x = 2 if both adults own a cell phone. The probabilities are
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
73
written in the table using the tree diagram above. The probability that x = 1 is obtained by adding the
probabilities of YN and NY
Outcomes
NN
YN or NY
YY
5.17
x
0
1
2
P(x)
.1296
.4608
.4096
Let: Y = teen said teachers were “totally clueless” about using the internet for teaching and learning.
N = teen said teachers were not “totally clueless” about using the internet for teaching and learning.
Then P(Y) = .78 and P(N) = 1 – .78 = .22
Let x denote the number of teens in a sample of two teens who believe that teachers are “totally
clueless” about using the internet for teaching and learning. The following table lists the probability
distribution of x.
Outcomes
NN
AN or NA
AA
5.19
Let
x
0
1
2
P(x)
.0484
.3432
.6084
A = first athlete selected used drugs
C = second athlete selected used drugs
B = first athlete selected did not use drugs
D = second athlete selected did not use drugs
Let x be the number of athletes who used illegal drugs in a sample of two athletes. The following table
lists the probability distribution of x.
Outcomes
BD
AD or BC
AC
x
0
1
2
P(x)
.4789
.4422
.0789
74
5.21
Chapter Five
a.
x
0
1
2
3
P(x)
.16
.27
.39
.18
xP(x)
0
.27
.78
.54
 xP( x)  1.59
x2P(x)
0
.27
1.56
1.62
 x 2 P(x)=3.45
   xP( x)  1.590
   x 2 Px    2 )  3.45  (1.59 ) 2  .960
TI-83: Insert numbers as two lists and then do the calculations as follows: press the 2 nd key, then
STAT, use the arrows to move over to MATH, use the arrows to move down to 5 sum or by pressing
the number 5 key, and press ENTER. To sum the lists such that we obtain Σ x* P(x) next press the
2nd key, the number 1 key, then typing the * key, next press the 2 nd key, the number 2 key, entering
the symbol ) followed by pressing the key for ENTER. Next we calculate Σx2 * P(x) by the previous
method only this time we insert a 2 before typing in the * key. Now we have the part necessary to
complete our calculation by following the formula for σ.
Sum (L1 * L2 )
1.590
Sum (L1 2 * L2 )
3.45
MINITAB: Enter the data for x and P(x) into columns C1 and C2 with the labels x and P(x) in the
gray row just below the column numbers. To calculate ΣxP(x) select Calc and then Calculator
causing a pop-up-box to open. Under the word “Expression” type in C1* C2 and in the box beside
“store result in variable” type C3. Label this column xP(x). Do the same for Σx2P(x) except type the
expression as C1*C1*C2 it is stored in C4, and labeled xxP(x). Then total the columns by selecting
Calc and then Column Statistics causing a pop-up-box to open. Under the word Statistic is Sum and
click on the box to the left of it and in the space beside the words Input variable type in the column
to be added up which in this case is xP(x) the first time and xxP(x) the second time through. Now we
have the parts necessary to complete our calculation by following the formula for σ.
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
75
Excel: Enter the data in two columns and at the top of each column be sure to label them. To
calculate the sum of the product of xP(x) in an empty cell type the formula =SUMPRODUCT(cell
range 1,cell range 2) and in this example it becomes =SUMPRODUCT(B2:B7,C2:C7). To get the
sum of m2f enter the formula =SUMPRODUCT(cell range 1, cell range 1, cell range 2) which in this
example is =SUMPRODUCT(B2:B7,B2:B7,C2:C7). Note: all of these formulas can be found by
selecting Insert, Function, and scrolling down until the function name appears in the box. Then all
you need to do is enter the appropriate cell ranges. Now we have the parts necessary to complete our
calculation by following the formula for σ.
b.
x
6
7
8
9
P(x)
.40
.26
.21
.13
xP(x)
2.40
1.82
1.68
1.17
 xP( x)  7.07
x2P(x)
14.40
12.74
13.44
10.53
 x 2 P( x)  51 .11
   xP( x)  7.070
   x 2 P( x)   2  51 .11  (7.07 ) 2  1.061
TI-83: Insert numbers as two lists and then do the calculations as follows: press the 2nd key, then
STAT, use the arrows to move over to MATH, use the arrows to move down to 5 sum or by pressing
the number 5 key, and press ENTER. To sum the lists such that we obtain Σ x* P(x) next press the
2nd key, the number 1 key, then typing the * key, next press the 2nd key, the number 2 key, entering
the symbol ) followed by pressing the key for ENTER. Next we calculate Σx 2 * P(x) by the previous
method only this time we insert a 2 before typing in the * key. Now we have the parts necessary to
complete our calculation by following the formula for σ.
Sum (L1 * L2 )
7.070
Sum (L1 2 * L2 )
51.11
MINITAB: Enter the data for x and P(x) into columns C1 and C2 with the labels x and P(x) in the
gray row just below the column numbers. To calculate ΣxP(x) select Calc and then Calculator
76
Chapter Five
causing a pop-up-box to open. Under the word “Expression” type in C1* C2 and in the box beside
“store result in variable” type C3. Label this column xP(x). Do the same for Σx2P(x) except type the
expression as C1*C1*C2 it is stored in C4, and labeled xxP(x). Then total the columns by selecting
Calc and then Column Statistics causing a pop-up-box to open. Under the word Statistic is Sum and
click on the box to the left of it and in the space beside the words Input variable type in the column
to be added up which in this case is xP(x) the first time and xxP(x) the second time through. Now we
have the parts necessary to complete our calculation by following the formula for σ.
Excel: Enter the data in two columns and at the top of each column be sure to label them. To
calculate the sum of the product of xP(x) in an empty cell type the formula =SUMPRODUCT(cell
range 1,cell range 2) and in this example it becomes =SUMPRODUCT(B2:B7,C2:C7). To get the
sum of m2f enter the formula =SUMPRODUCT(cell range 1, cell range 1, cell range 2) which in this
example is =SUMPRODUCT(B2:B7,B2:B7,C2:C7). Note: all of these formulas can be found by
selecting Insert, Function, and scrolling down until the function name appears in the box. Then all
you need to do is enter the appropriate cell ranges. Now we have the parts necessary to complete our
calculation by following the formula for σ.
5.23
x
0
1
2
3
4
P(x)
.73
.16
.06
.04
.01
xP(x)
0
.16
.12
.12
.04
 xP(x)  .44
   xP ( x)  .440 error
   x 2 P( x)   2  .92  (.44)2  .852 error
x2 P(x)
.00
.16
.24
.36
.16
2
 x P( x)  .92
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
5.25
x
0
1
2
3
4
5
6
P(x)
.05
.12
.19
.30
.20
.10
.04
xP(x)
0
.12
.38
.90
.80
.50
.24
 xP(x)  2.94
x2 P(x)
0
.12
.76
2.70
3.20
2.50
1.44
2
 x P( x)  10 .72
   xP ( x)  2.94 camcorders
   x 2 P( x)   2  10.72  (2.94)2  1.441 camcorders
On average, 2.94 camcorders are sold per day at this store.
5.27
x
0
1
2
P(x)
.25
.50
.25
xP(x)
0
.50
.50
xP
(x
)  1.00

x2 P(x)
0
.50
1.00
2
x
P
( x)  1.50

   xP ( x)  1.000 head
   x 2 P( x)   2  1.50  (1.00 ) 2  .707 heads
The value of the mean,  = 1.00, indicates that, on average, we will expect to obtain 1 head in every
two tosses of the coin.
5.29
x
0
1
2
3
4
P(x)
.048
.388
.292
.164
.108
xP(x)
0
.388
.584
.492
.432
 xP(x)  1.896
x2 P(x)
0
.388
1.168
1.476
1.728
2
 x P( x)  4.760
   xP( x)  1.896  1.9 TV sets
   x 2 P( x)   2  4.76  (1.896 ) 2  1.079 TV sets
Thus, there is an average of 1.90 TV sets per family in this town, with a standard deviation of 1.079
sets.
77
78
Chapter Five
5.31
x
0
1
2
P(x)
.9025
.0950
.0025
xP(x)
0
.0950
.0050
xP
 ( x)  .1000
x2 P(x)
0
.0950
.0100
2
 x P( x)  .1050
   xP( x)  .10 car
   x 2 P( x)   2  .1050  (.10 ) 2  .308 car
5.33
x
10
5
2
0
P(x)
.15
.30
.45
.10
xP(x)
1.50
1.50
0.90
0
xP
(x
)  3.9

x2 P(x)
15.00
7.50
1.80
0
2
x
P
(
x
)  24.30

   xP( x)  $3.9 million
   x 2 P( x)   2  24 .3  (3.9)2  $3.015 million.
Thus, the contractor is expected to make $3.9 million profit with a standard deviation of $3.015
million.
5.35
x
0
1
2
P(x)
.5455
.4090
.0455
xP(x)
0
.4090
.0910
 xP(x)  .5000
x2 P(x)
0
.4090
.1820
2
 x P( x)  .5910
   xP( x)  .500 person
   x 2 P( x)   2  .5910  (.50 ) 2  .584 person
5.37
3! = 3  2  1 = 6
(9 -3)! = 6! = 6  5  4  3  2  1 = 720
9! = 9  8  7  6  5  4  3  2  1 = 362,880
(14 – 12)! = 2! = 2  1 = 2
5 C3 =
7 C4
=
9 C3 =
5!
5!
5  4  3  2 1
=
=
= 10
5  3 !  3 ! 2 !  3 ! 2  1 3  2  1
7!
7!
7  6  5  4  3  2 1
=
=
= 35
7  4 !  4 ! 3 !  4 ! 3  2  1 4  3  2  1
9!
9!
=
=
9  3 !  3 ! 6 !  3 !
9  8  7  6  5  4  3  2 1
= 84
6  5  4  3  2 1 3  2 1
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
4 C0 =
4  3  2 1
4!
4!
=
=
=1
4  0 ! 0 ! 4 !  0 ! 4  3  2 11
3 C3 =
3!
3!
3  2 1
=
=
=1
3  3 !  3 ! 0 !  3 ! 1  3  2  1
79
TI-83: To enter a factorial, first enter the number then pressing the MATH button, highlighting PRB,
pressing number 4, and then pressing ENTER. In this example its 3 then MATH, PRB, 4, and ENTER.
3!
(14-12)!
6
2
(9-3)!
720
9!
362880
To enter a combination function, first enter the number of elements or the n, then pressing the MATH
button, highlighting PRB, pressing number 3, entering the second number called x (or labeled r by the
calculator) and then pressing ENTER. In the first example of a combination 5 C3 its 5 then MATH,
PRB, the formula 3, the number 3 and ENTER.
5 nCr 3
4 nCr 0
10
7 nCr 4
1
3 nCr 3
35
1
9 nCr 3
84
MINITAB: Will do factorials but will not do combinations. Using the factorial function will give you
the three parts of the combination formula so you can complete the rest on your own. To do the
factorials select CALC, Calculator, under Functions scroll down and highlight Gamma, and select a
destination on the spread sheet for our answer by typing the cell number beside the words store result
in variable. Next below the word Expression type in a number that is one number larger than the
factorial you desire. In the first example we want 3! so we type in the number 4, making sure we have
already entered a cell for our result and then click OK.
80
Chapter Five
Excel: In Excel the factorial function is =FACT(number) and the combination function is
=COMBIN(number of elements, number in selection). Also be aware that we can also place an
equation in place of the number parenthesis. For the first example of a factorial we select an empty cell
and type =FACT(3). We can then repeat this for all of the other factorials in this problem. For the first
combination, we select another empty cell and type =COMBIN(5, 3); repeating this for all of the other
combinations.
5.39
The total number of ways to select two faculty members from 16 is:
16 C 2 =
5.41
16!
16  15  14  13  12  11  10  9  8  7  6  5  4  3  2  1
=
= 120
16  2 !  2 ! 14  13  12  11  10  9  8  7  6  5  4  3  2  1 2  1
Total possible selections for selecting three horses from 12 are:
12 C 3 =
12  11  10  9  8  7  6  5  4  3  2  1
12!
=
= 220
12  3 !  3 ! 9  8  7  6  5  4  3  2  1  3  2  1
20!
=38,760
(20  6) !  6 !
5.43
Total number of ways of selecting six stocks from 20 are: 20 C 6 =
5.45
Total number of ways of selecting 9 items from a population of 20 are: 20 C 9 =
20!
= 167,960
(20  9) !  9 !
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
5.47
81
a. An experiment that satisfies the following four conditions is called a binomial experiment:
i. There are n identical trials. In other words, the given experiment is repeated n times. All these
repetitions are performed under similar conditions.
ii. Each trial has two and only two outcomes. These outcomes are usually called a success and a
failure.
iii. The probability of success is denoted by p and that of failure by q, and p + q =1. The probability
of p and q remain constant for each trial.
iv. The trials are independent. In other words, the outcome of one trial does not affect the outcome
of another trial.
b. Each repetition of a binomial experiment is called a trial.
c. A binomial random variable x represents the number of successes in n independent trials of a
binomial experiment.
5.49
a. This is not a binomial experiment because there are more than two outcomes for each repetition.
b. This is an example of a binomial experiment because it satisfies all four conditions of a binomial
experiment:
i.
There are many identical rolls of the die.
ii.
Each trial has two outcomes: an even number and an odd number.
iii. The probability of obtaining an even number is ½ and that of an odd number is ½. These
probabilities add up to 1, and they remain constant for all trials.
iv. All rolls of the die are independent.
c. This is an example of a binomial experiment because it satisfies all four conditions of a binomial
experiment:
i
There are many identical trials (selection of voters).
ii.
Each trial has two outcomes: a voter favors the proposition and a voter does not favor the
proposition.
iii. The probability of the two outcomes are .54 and .46 respectively. These probabilities add up
to 1. These two probabilities remain the same for all selections.
iv. All voters are independent.
5.51
a. n = 8, x = 5, n – x = 8 – 5 = 3, p = .70, and
q = 1 – p = 1 – .70 = .30
P(x = 5) = n C x p x q n x = 8 C 5 (.70) 5 (.30) 3 = (56) (.16807) (.027) = .2541
b. n = 4, x = 3, n – x = 4 – 3 = 1, p = .40, and
q = 1 – p = 1 – .40 = .60
P(x = 3) = n C x p x q n x = 4 C 3 (.40) 3 (.60) 1 = (4) (.064) (.60) = .1536
82
Chapter Five
c. n = 6, x = 2, n – x = 6 – 2 = 4, p = .30, and
q = 1 – p = 1 – .30 = .70
P(x = 2) = n C x p x q n x = 6 C 2 (.30) 2 (.70) 4 = (15) (.09) (.2401) = .3241
TI-83: Press 2nd and VARS, then scroll down to the tenth option which is listed as 0:binompdf( and
press ENTER. Now enter three values one for n, one for p and the last one for x. Be sure to separate
them with a comma and then press enter. For part a, we type in 8,0.7, 5) and press ENTER. Here ) is
optional as the command will work without it.
binompdf(8, .7, 5)
.25412184
binompdf(4, .4, 3)
.1536
binompdf(6, .3, 2)
.324135
MINITAB: Select CALC, Probability Distribution, and then Binomial. In the pop-up-box, click next
to Probability and Input Constant. Beside the words Number of Trials insert your n, beside the words
probability of success insert you p, beside the words Input Constant insert your x, and click on OK.
The answer is then displayed in the session window.
Excel: In an empty cell type in the formula =BINOMDIST(x, n, p, cumulative). Where x is the number
of successes, n is the number of trials and p is the probability. Use FALSE as the value for cumulative.
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
a.
x
0
1
2
3
4
5
6
7
P(x)
.0824
.2471
.3177
.2269
.0972
.0250
.0036
.0002
0.4
0.3
P(x)
5.53
83
0.2
0.1
0
0
1
2
3
4
5
6
7
x
b.   np  (7)(.30)  2.100
  npq  (7)(. 30 )(. 70 )  1.212
T1-83, MINITAB, and Excel: Use the procedure described in detail in 5.51 for each of the values of
x. The Graph the bar graph as described in Chapter 2.
Note for TI-83: For the TI-83 follow the procedure in 5.51 but instead of entering your x after p
instead press the ) key and press ENTER. In the row below you will get all of the probabilities of x if
you scroll right using your right arrow key.
Note for MINITAB: A faster way to accomplish this is to type in all the values of x in C1, then Select
CALC, Probability Distribution, and then Binomial. In the pop-up-box, click next to Probability and
Input Column. Beside the words Number of Trials insert your n, beside the words probability of
success insert you p, beside the words Input Column insert, beside the words Optional Storage type
C2, and click on OK. The answer is then displayed on the worksheet. The appropriate labels have been
added for clarity.
84
5.55
Chapter Five
i. Let n = 5 and p = .50. The probability distribution and probability graph for this case are shown
below. As we can observe, the probability distribution is symmetric in this case.
0.4
x
P(x)
0
.0312
1
.1562
2
.3125
3
.3125
4
.1562
5
.0312
ii. Let n = 5 and p = .20. The
P(x)
0.3
0.2
0.1
0
0
1
2
3
4
5
x
distribution and
probability
probability
graph for this case are shown below. As we can observe, the probability distribution is skewed to
the right in this case.
0.5
P(x)
.3277
.4096
.2048
.0512
.0064
.0003
0.4
P(x)
x
0
1
2
3
4
5
0.3
0.2
0.1
0
0
iii. Let n = 5 and p = .70. The
1
2
3
4
5
x
probability distribution and probability graph for this case are shown below. As we can observe,
the probability distribution is skewed to the left in this case.
0.4
P(x)
.0024
.0284
.1323
.3087
.3601
.1681
0.3
P(x)
x
0
1
2
3
4
5
0.2
0.1
0
0
5.57
1
2
3
4
5
x
a. Here, n = 10 and p = .24
The random variable x can assume any of the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10.
b. P(x = 4) = n C x p x q n x =
5.59
10 C 4
(.24)4 (.76)6 = (210) (.00331776) (.192699928)= .1343
Here, n = 15 and p = .80
Let x denote the number of adults in a random sample of 15 who feel stress “frequently” or
“sometimes” in their daily lives.
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
85
a. P(at most 9) = P(x≤ 9) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9)=
=.0000 + .0000 + .0000 = .0000 + .0001 + .0007 + .0035 + .0138+ .0430 = .0611
b. P(at least 11) = P(x ≥ 11) = P(11) + P(12) + P(13) + P(14) + P(15)
= .1876 + .2501 + .2309 + .1319 + .0352 = .8357
c. P(10 ≤ x ≤ 12) = + P(10) + P(11) + P(12) = .1032 + .1876 + .2501 = .5409
5.61
Here, n = 10 and p = .349
a. P(exactly 4) = P(4) = n C x p x q n  x 10 C 4 (.349)4(.651)6 =(210) (.014835483) (.076117748) = .2371
b. P(none) = P(0) = n C x p x q n  x 10 C 0 (.349) 0 (.651)10  (1) (1) (.013671302) = .0137
c. P(exactly 8) = P(8) = n C x p x q n  x 10 C8 (.349) 8 (.651) 2  (45) (.00022009) (.423801) = .0042
5.63
Here, n = 8 and p = .85
a. P(exactly 8) = P(8) = n C x p x q n  x  8 C8 (.85) 8 (.15) 0  (1) (.27249053) (1) = .2725
b. P(exactly 5) = P(5) = n C x p x q n  x  8 C5 (.85) 5 (.15) 3  (56) (.4437053) (.003375) = .0839
Here, n = 7 and p = .80
a.
x
0
1
2
3
4
5
6
7
P(x)
.0000
.0004
.0043
.0287
.1147
.2753
.3670
.2097
0.4
0.3
P(x)
5.65
0.2
0.1
0
The mean and standard deviation of x are:
  np  (7)(.80)  5.6 customers
  npq  (7)(.80)(.20)  1.058 customers
b. P(exactly 4) = P(4) = .1147
0
1
2
3
4
5
6
7
x
86
a. Here, n = 8 and p = .70
x
0
1
2
3
4
5
6
7
8
P(x)
.0001
.0012
.0100
.0467
.1361
.2541
.2965
.1977
.0576
0.3
0.2
P(x)
5.67
Chapter Five
0.1
0
0
1
2
3
4
5
6
7
8
x
The mean and standard deviation of x are:
  np  (8)(.70)  5.600 customers
  npq  (8)(.70)(.30)  1.296 customers
b. P(exactly 3 customers like the hamburger) = P(3) = .0467
5.69
a. P x  2 
r C x N r C n x
N
b. P x  0 
Cn
r C x N r C n x
N
Cn


3 C 2 8 3 C 4  2
8 C4
3 C 0 8 3 C 4  0
c. P x  1  P0  P1  .0714 
8 C4
 310  / 70  .4286
 15 / 70  .0714
3 C1 83 C 41
8 C4
 .0714  310  / 70  .0714  .4286  .5000
TI-83: It does not have a built in function for the hypergeometric distribution. Instead use the nCr
function used earlier in Chapter 5 and presented here in question 5.37.
MINITAB: Select CALC, Probability Distribution, and then Hypergeometric. In the pop-up-box, click
next to Probability and Input Constant. Beside the words Population size insert your N, beside the
words Success in Population insert you r, beside the words sample size insert your n, beside the words
Input Constant insert your x, and click on OK. For part a of this question N is 8, r is 3, n is 4, and x is
2. The answer is then displayed in the session window.
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
87
Note: Just like with the binomial distribution in 5.53 you can generate the probabilities of all x at once
by typing all the values into C1 first and the following the procedure above but instead of clicking
beside Input Constant this time click beside Input Column. Beside Input Column put C1 and beside
Optional Storage put C2. Labels for C1 and C2 have been added for easier reading. The results appear
in the worksheet and are shown below. Note for part c remember to add P(x=0) and P(x=1) to get P
(x  1).
Excel: In an empty cell type in the formula =HYPGEOMDIST(x, n, r, N). Where x is the number of
successes in the trials, n is the number of trials, r is the number of successes in the population, and N is
the size of the population.
5.71
a. P x  2 
r C x N r C n x
N
b. P x  4 
Cn
r C x N r C n x
N
Cn
c. P x  1  P0  P1 


4 C 2 11 4 C 4  2
11 C 4
4 C 4 11 4 C 4  4
11 C 4
4 C 0 11 4 C 4 0
11 C 4
 .1061  .4242  .5303

 621 / 330  .3818
 11 / 330  .0030
4 C1 11 4 C 41
11 C 4
 135  / 330  435  / 330
88
5.73
Chapter Five
Let x be the number of corporations that incurred losses in a random sample of 3 corporations, and r be
the number of corporations in 15 that incurred losses. Then,
a. P(exactly 2) = P x  2 
b. P(none) = P x  0 
9 C2 6 C1
15 C3
9 C0 6 C3
15 C 3
N = 15, r = 9, N – r = 6, and n = 3.
 36 6 / 455  .4747
 120  / 455  .0440
c. P(at most 1) = P x  1  P0  P1 
9 C0 6 C3
15 C 3

9 C1 6 C 2
15 C 3
 120  / 455  915  / 455
 .0440  .2967  .3407
5.75
Let x be the number of extremely violent games in a random sample of three, and r be the number of
extremely violent games in eleven. Then, N = 11, r = 4, and N - r = 7, and n = 3.
a. P(x = 2) =
r C x N r C n x
N
Cn
=
4 C 2 11 4 C 3 2
11 C 3
b. P(x > 1) = P(2) + P(3) = .2545 +
c. P(x = 0) =
5.77
r C x N r C n x
N Cn
=
=
6(7 )
=.2545
165
4 C 3 11 4 C 33
11 C 3
4 C 0 11 4 C 30
11 C 3
=
= .2545 +
4(1)
= .2545 + .0242 = .2787
165
1(35 )
= .2121
165
The following three conditions must be satisfied to apply the Poisson probability distribution:
i. x is a discrete random variable.
ii. The occurrences are random, that is, they do not follow any pattern.
iii. The occurrences are independent.
5.79
a. P(x  1) = P(x = 0) + P(x = 1) =
(1) (.00673795 ) (5) (.00673795 )
(5) 0 e 5 (5)1 e 5



1
1
0!
1!
= .0067 + .0337 = .0404
Note that the value of e-5 is obtained from Table V of Appendix C of the text.
b. P(x = 2) =
2.52 e 2.5
2!

6.25 .08208500   .2565
2
TI-83: Press 2nd and VARS, then scroll down to the tenth option which is listed as B: poissonpdf( or C:
poissoncdf ( and press ENTER. Use B for the probability of a single x and C when you want the
cumulative probability that x is less to or equal to some number. Now enter the two values one for 
and the other for x. Be sure to separate them with a comma and then press enter. For part a, we use the
formula C: poissoncdf ( and type in 5,1) and press ENTER. Here ) is optional as the command will
work without it. For part b, we use we use the formula B: poissonpdf ( and type in 2.5,2) and press
ENTER. The results are shown below.
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
89
poissoncdf(5, 1)
.040427682
poissonpdf(2.5,.2)
..2565156207
MINITAB: Select CALC, Probability Distribution, and then Poisson. For part a in the pop-up-box,
click next to Cumulative Probability and Input Constant. The will generate the probability that x is less
than or equal to a number which in this example is 1. Beside the word Mean type in your  and beside
the words Input Constant insert your x, and click on OK. For part a of this question  is 5 and x is less
than or equal to 1. The answer is then displayed in the session window.
For part b in the pop-up-box, click next to Probability and Input Constant. The will generate the
probability that x is equal to a number which in this example is 2. Beside the word Mean type in your
 and beside the words Input Constant insert your x, and click on OK. For part b of this question  is
2.5 and x is equal to 2. The result is displayed in the session window.
Note: Just like with the binomial distribution in 5.53 you can generate the probabilities of all x at once
by typing all the values into C1 first and the following the procedure above but instead of clicking
beside Input Constant this time click beside Input Column. Beside Input Column put C1 and beside
Optional Storage put C2. This is true of for both probabilities and cumulative probabilities.
Excel: In an empty cell type in the formula =POISSON(x, mean, cummulative) and then press
ENTER. Where x is the number of successes in the trials, mean is  , and cumulative describes if this
is a cumulative probability or not. For part a, x is less than or equal to 1,  is 5, and cumulative is true
as we are essentially look for the probability that x is 1 and the probability that is is 0. For part b, x is
equal to 2,  is 2.5, and cumulative is false as we are looking for the probability of only one particular
x. The results are shown below where labels were added for easier reading.
90
a. Probability distribution of x for  = 1.3
x
0
1
2
3
4
5
6
7
8
P(x)
.2725
.3543
.2303
.0998
.0324
.0084
.0018
.0003
.0001
0.4
0.3
P(x)
5.81
Chapter Five
0.2
0.1
0
0
1
2
3
4
5
6
7
7
8
8
x
The mean, variance, and standard deviation are:
 2    1.3, and     1.3  1.140
    1.3,
b. Probability distribution of x for  = 2.1
P(x)
.1225
.2572
.2700
.1890
.0992
.0417
.0146
.0044
.0011
.0003
.0001
0.3
0.2
P(x)
x
0
1
2
3
4
5
6
7
8
9
10
0.1
0
0
1
2
The mean, variance, and standard deviation are:
 2    2.1, and     2.1  1.449
    2.1,
5.83
 = 1.7 pieces of junk mail per day and x = 3
P( x  3) 
5.85
 x e 
x!

(1.7) 3 e 1.7 4.913 18268352 

 .1496
3!
6
 = 5.4 shoplifting incidents per day and x = 3
P( x  3) 
 x e 
x!

(5.4) 3 e 5.4 157 .464 .00451658 

 .1185
3!
6
3
4
5
6
9 10
x
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
5.87
91
 = 3.7 reports of lost students’ ID cards per day
a. P(at most 1) = P(0) + P(1)=
3.70 e 3.7
0!

(3.7)1 e 3.7 1.02472353  (3.7) (.02472353 )


1!
1
1
= .0247 + .0915 = .1162
b. i. P(1 to 4) = P(1) + P(2) + P(3) + P(4) = .0915 + .1692 + .2087 + .1931 = .6625
ii. P(at least 6) = P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12) + P(13)
= .0881 + .0466 + .0215 + .0089 + .0033 + .0011 + .0003 + .0001 = .1699
iii. P(at most 3) = P(0) + P(1) + P(2) + P(3) = .0247 + .0915 + .1692 + .2087 = .4941
5.89
 = .5 defect per 500 yards
a. P( x  1) 
 x e 
x!

(.5)1 e .5 .5.60653066 

 .3033
1!
1
b. i. P(2 to 4) = P(2) + P(3) + P(4) = .0758 + .0126 + .0016 = .0900
ii. P(more than 3) = P(4) + P(5) + P(6) + P(7) = .0016 + .0002 + .0000 + .0000 = .0018
iii. P(less than 2) = P(0) + P(1) = .6065 + .3033 = .9098
5.91
Let x be the number of customers that come to this savings and loan during a given hour. Since the
average number of customers per half hour is 4.8, the average number per hour is 2 × 4.8 = 9.6. Thus,
 = 9.6.
a. P(exactly 2) = P( x  2) 
 x e 
x!

(9.6) 2 e 9.6 92 .16 .00006773 

 .0031
2!
2
b. i. P(2 or less) = P(x  2) = P(0) + P(1) + P(2) = .0001 + .0007 + .0031 = .0039
ii. P(10 or more) = P(x  10) = P(10) + P(11) + P(12) +… P(24)= .1241 + .1083 + .0866 + .0640 +
.0439 + .0281 + .0168 + .0095 + .0051+ .0026 + .0012 + .0006 + .0002 + .0001 + .0000 = .4911
5.93
Let x be the number of policies sold by this salesperson on a given day. Since the salesperson sells an
average of 1.4 policies per day,  = 1.4.
a. P(none) = P( x  0) 
 x e 
x!

(1.4) 0 e 1.4 1.24659696 

 .2466
0!
1
92
Chapter Five
b.
x
0
1
2
3
4
5
6
7
8
P(x)
.2466
.3452
.2417
.1128
.0395
.0111
.0026
.0005
.0001
c. The mean, variance, and standard deviation are:
    1.4,
5.95
 2    1.4, and     1.4  1.183
Let x denote the number of households in a random sample of 50 who own answering machines.
Since, on average, 20 households in 50 own answering machines,  = 20.
a. P(exactly 25 own answering machines) = P( x  25) 
 x e 
x!

(20 ) 25 e 20
 .0446
25!
b. i. P(at most 12 own answering machines) = P(x  12) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)
+ P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12) = .0000 + .0000 + .0000 + .0000 + .0000
+ .0001 + .0002 + .0005 + .0013 + .0029 + .0058 + .0106 + .0176 = .0390
ii. P(13 to 17) = P(13  x  17) = P(13) + P(14) + P(15) + P(16) + P(17)
= .0271 + .0387 + .0516 + .0646 + .0760 = .2580
iii. P(at least 30 own answering machines) = P(x  30)
= P(30) + P(31) + P(32) + … + P(39)
= .0083 + .0054 + .0034 + .0020 + .0012 + .0007 + .0004 + .0002 + .0001 + .0001 = .0218
5.97
x
2
3
4
5
6
P(x)
.05
.22
.40
.23
.10
xP(x)
.10
.66
1.60
1.15
.60
 xPx  4.11
   xPx  = 4.11 cars
   x 2 Px    2  17 .93  4.112  1.019 cars
This mechanic repairs, on average, 4.11 cars per day
x2P(x)
.20
1.98
6.40
5.75
3.60
 x 2 Px  17.93
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
5.99
93
a. & b.
x
-1.2
-.7
.9
2.3
P(x)
.17
.21
.37
.25
X2 P(x)
.2448
.1029
.2997
1.3225
xP(x)
-.204
-.147
.333
.575
 xPx   .557
 x 2 Px  1.9699
   xPx  = $.557 million = $557,000
   x 2 Px    2  1.9699  .557 2  $1.288274 million = $1,288,274
The value of  = $557,000 indicates that the company has an expected profit of $557,000 for next
year.
5.101
Let x denote the number of machines that are broken down at a given time. Assuming machines are
independent, x is a binomial random variable with n = 8 and p = .04.
a. P(all 8 are broken down) = P(x = 8) = n C x p x q n x  8 C8 0.04 8 .96 0
 1.0000000000 07 1  .0000
b. P(exactly 2 are broken down) = P(x = 2) = n C x p x q n  x  8 C 2 0.04 2 .96 6
 28 .0016 .78275779   .0351
c. P(none is broken down) = P(x = 0) = n C x p x q n x  8 C 0 0.04 0 .96 8  11.72138958   .7214
5.103
Let x denote the number of defective motors in a random sample of 20. Then x is a binomial random
variable with n = 20 and p = .05.
a. P(shipment accepted) = P(x  2) = P(0) + P(1) + P(2) = .3585 + .3774 + .1887 = .9246
b. P(shipment rejected) = 1 – P(shipment accepted) = 1 - .9246 = .0754
5.105
Let x denote the number of households who own homes in the random sample of 4 households. Then x
is a hypergeometric random variable with N = 15, r = 9, and n = 4.
a. P(exactly 3) = P(x = 3) =
r C x N r C n x
N
Cn
b. P(at most 1) = P( x  1)  P(0) P(1) 

9 C 3 6 C1
15 C 4
9 C0 6 C4
15 C 4


(84 )( 6)
 .3692
1365
9 C1 6 C 3
15 C 4

(1)(15 ) (9)( 20 )

1365
1365
 .0110 .1319  .1429
c. P(exactly 4) = P(x = 4) =
r C x N r C n x
N Cn

9 C4 6 C0
15 C 4

(126 )(1)
 .0923
1365
94
5.107
Chapter Five
Let x denote the number of defective parts in a random sample of 4. Then x is a hypergeometric
random variable with N = 16, r = 3, and n = 4.
a. P(shipment accepted) = P( x  1)  P(0) P(1) 

3 C 0 13 C 4
16 C 4

3 C1 13 C 3
16 C 4
(1)( 715 ) (3)( 286 )

 .3929 .4714  .8643
1820
1820
b. P(shipment not accepted) = 1 – P(shipment accepted) = 1 – .8643 = .1357
5.109
Here,  = 7 cases per day
a. P(x = 4) =
 x e 
x!

(7) 4 e 7 2401 .00091188 

 .0912
4!
24
b. i. P(at least 7) = P(7) + P(8) + . . . + P(18)= .1490 + .1304 + .1014 + .0710 + .0452 + .0263 +
.0142 + .0071 + .0033 + .0014 + .0006 + .0002 + .0001 = .5502
ii. P(at most 3) = P(0) + P(1) + P(2) + P(3)= .0009 + .0064 + .0223 + .0521 = .0817
iii. P(2 to 5) = P(2) + P(3) + P(4) + P(5)= .0223 + .0521 + .0912 + .1277 = .2933
5.111
Here,  = 1.4 airplanes per hour
a. P(x = 0) =
 x e 
x!

(1.4) 0 e 1.4 1.24659696 

 .2466
0!
1
b.
x
0
1
2
3
4
5
6
7
8
5.113
P(x)
.2466
.3452
.2417
.1128
.0395
.0111
.0026
.0005
.0001
Let x be a random variable that denotes the gain you have from this game. The probability for each
number is not the same, however. There are 36 different outcomes for two dice: (1,1), (1,2), (1,3),
(1,4), (1,5), (1.6), (2,1), (2,2),…, (6,6).
1
P(sum = 2) = P(sum = 12) =
36
P(sum = 3) = P(sum = 11) =
2
36
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
P(sum = 4) = P(sum = 10) =
P(sum = 9) =
95
3
36
4
36
P(x = 20) = P(you win) = P(2) + P(3) + P(4) + P(9) + P(10) + P(11) + P(12)

1
2
3
4
3
2
1 16







36 36 36 36 36 36 36 36
P(x = -20) = P(you lose) = 1 – P(you win) = 1 –
x
P(x)
16
 .4444
36
20
 .5556
36
20
-20
16 20

36 36
xP(x)
8.89
-11.11
 xP(x)  2.22
The value of
 xP(x) = -2.22 indicates that your expected “gain” is -$2.22, so you should not accept
this offer. This game is not fair to you since you are expected to lose $2.22.
5.115
a. Team A needs to win four games in a row, each with probability .5, so P(team A wins the series in
four games) = .54 = .0625
In order to win in five games, Team A needs to win 3 of the first four games as well as the fifth
game, so P(team A wins the series in five games) = 4 C3 .53 .5.5  .125 .
b. If seven games are required for a team to win the series, then each team needs to win three of the
first six games, so P(seven games are required to win the series) =
5.117
6 C3
.53 .53  .3125 .
a. Let x denote the number of drug deals on this street on a given night. Note that x is discrete. This
text has covered two discrete distributions, the binomial and the Poisson. The binomial distribution
does not apply here, since there is no fixed number of “trials”. However, the Poisson distribution
might be appropriate.
b. To use the Poisson distribution we would have to assume that the drug deals occur randomly and
independently.
c. The mean number of drug deals per night is three, so for the Poisson distribution for one night,
 = 3. If the residents tape for two nights, then  = 2 x 3 = 6.
Thus, P(film at least 5 drug deals) = P(x  5) = 1 – P(x < 5)
96
Chapter Five
= 1 – [P(0) + P(1) + P(2) + P(3) + P(4)] = 1 – (.0025 + .0149 + .0446 + .0892 + .1339) = .7149
d. Part c. shows that two nights of taping are insufficient, since P(x  5) = .7149 < .90. Try taping
for three nights. Then  = 3 x 3 = 9. P(x  5) = 1 – (.0001 + .0011 + .0050 + .0150 + .0337) =
.9451. This exceeds the required probability of .90, so the camera should be rented for three nights.
5.119
Let  be the mean number of cheesecakes sold per day. Here  =5. Let x be the number of sales per
day. We want to find k such that P(x > k) < .1. Using the Poisson probability distribution we find that
P(x > 7) = 1 – P (x  7) = 1 - .867 = .133 and P(x > 8) = 1 – P(x  8) = 1 - .932 = .068. So, if the
baker wants the probability of losing a sale to be less than .1, he needs to make 8 cheesecakes.
5.121
a. There are 7 C 4 = 35 ways to choose four questions from the set of seven.
b. The teacher must choose both questions that the student did not study ( 2 C 2 ways to do this), and
any two of the remaining five questions ( 5 C 2 ways to do this).
Thus, there are 2 C 2 5 C 2 = (1) (10) = 10 ways to choose four questions that include the two that the
student did not study.
c. From the answers to parts a and b, P(the four questions on the test include both questions that the
student did not study) = 10/35 = .2857.
5.123
For each game, let x = amount you win
Game I:
Outcome
Head
Tail
x
3
-1
P(x)
.50
.50
xP(x)
1.50
-.50
 xP(x)  1.00
X2 P(x)
4.50
.50
 x 2 P( x)  5.00
   xP(x)  $1.00
   x 2 P( x)   2  5  (1) 2  $2.00
Game II:
Outcome
First ticket
Second ticket
Neither
   xP(x)  $. 90
x
300
150
0
P(x)
1/500
1/500
498/500
xP(x)
.60
.30
.00
xP
 (x)  .90
X2 P(x)
180
45
0
 x 2 P( x)  225
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
97
   x 2 P( x)   2  225  (.90) 2  $14 .97
Game III:
Outcome
Head
Tail
x
1,000,002
-1,000,000
P(x)
.50
.50
xP(x)
500,001
-500,000
xP
 (x)  1.00
X2 P(x)
5 x 1011
5 x 1011
 x 2 P(x)  1012
   xP(x)  $1.00
   x 2 P( x)   2  10 12  (1) 2  $1,000 ,000
Game I is preferable to Game II because the mean for Game I is greater than the mean for Game II.
Although the mean for Game III is the same as Game I, the standard deviation for Game III is
extremely high, making it very unattractive to a risk-adverse person. Thus, for most people, Game I is
the best and, probably, Game III is the worst (due to its very high standard deviation).
5.125
Let:
x1 = the number of contacts on the first day
x2 = the number of contacts on the second day
The following table, which may be constructed with the help of a tree diagram, lists the various
combinations of contacts during the two days and their probabilities. Note that the probability of each
combination is obtained by multiplying the probabilities of the two events included in that
combination.
x1 , x2
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(2, 1)
(2, 2)
(2, 3)
(2. 4)
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
Probability
.0144
.0300
.0672
.0084
.0300
.0625
.1400
.0175
.0672
.1400
.3136
.0392
.0084
.0175
.0392
.0049
y
2
3
4
5
3
4
5
6
4
5
6
7
5
6
7
8
The following table gives the probability distribution of y. This table is prepared from the previous table.
98
Chapter Five
y
2
3
4
5
6
7
8
P(y)
.0144
.0600
.1969
.2968
.3486
.0784
.0049
Self-Review Test for Chapter Five
1.
Random variable: A variable whose value is determined by the outcome of a random experiment is
called a random variable.
Discrete random variable: A random variable whose values are countable is called a discrete random
variable. An example of this is the number of students in a class.
Continuous random variable: A random variable that can assume any value in one or more intervals is
called a continuous random variable. An example of this is the height of a person.
2.
The probability distribution table.
3. a
6.
Following are the four conditions of a binomial experiment.
4. b
5. a
i. There are n identical trials. In other words, the given experiment is repeated n times. All these
repetitions are performed under similar conditions.
ii. Each trial has two and only two outcomes. These outcomes are usually called a success and a failure.
iii. The probability of success is denoted by p and that of failure by q, and p + q = 1. The probabilities p
and q remain constant for each trial.
iv. The trials are independent.
Example 5-16 in the text can be considered as an example of a binomial experiment.
7. b
8. a
9. b
10. a
11. c
13. a
12. A hypergeometric probability distribution is used to find probabilities for the number of successes in a
fixed number of trials, when the trials are not independent (such as sampling without replacement from a
small population.) Example: Select 2 balls without replacement from an urn that contains 3 red balls and 5
black balls. The number of red balls in the sample is a hypergeometric random variable.
14. Following are the three conditions that must be satisfied to apply the Poisson probability distribution.
i. x is a discrete random variable.
ii. The occurrences are random, that is, they do not follow any pattern.
iii. The occurrences are independent, that is, the occurrence (or nonoccurrence) of an event does not
influence the successive occurrences (or nonoccurrences) of that event.
Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual
99
15.
x
0
1
2
3
4
5
P(x)
.15
.24
.29
.14
.10
.08
xP(x)
.00
.24
.58
.42
.40
.40

xP
 x  2.04
X2 P(x)
.00
.24
1.16
1.26
1.60
2.00
 x 2 Px  6.26
   xP(x)  2.04 homes
   x 2 P( x)   2  6.26  (2.04 ) 2  1.449 homes
The four real estate agents sell an average of 2.04 homes per week.
16.
Here, n = 12 and p = .60
a. i. P(exactly 4) = P(4) = n C x p x q n x 12 C8 (.60) 8 (.40) 4  (495) (.01679616) (.0256) = .2128
ii. P(at least 6) = P(x  6) = P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12)
= .1766 + .2270 + .2128+ .1419 + .0639 + .0174 + .0022 = .8418
iii. P(less than 4) = P(x < 4) = P(0) + P(1) + P(2) + P(3) = .0000 + .0003 + .0025 + .0125 = .0153
b. μ = np = 12(.60) = 7.2 adults
and σ =
npq = 12 (.60 )(. 40 ) =1.697 adults
17. Let x denote the number of females in a sample of 4 volunteers from the 12 nominees. Then x is a
hypergeometric random variable with: N = 12, r = 8, N – r = 4 and n = 4.
a. P(x = 3) =
r C x N r C n x
N
b. P(x = 1) =
Cn
r C x N r C n x
N
Cn
c. P(x ≤ 1) = P(0) + P(1) =
=
=
8 C 3 128 C 4 3
12 C 4
8 C1 128 C 4 1
12 C 4
8 C 0 128 C 4  0
12 C 4
=
=
56 (4)
=.4525
495
8(4)
= .0646
495
+ .0646 =
1(1)
+ .0646 = .0020 + .0646 = .0666
495
18. Here,  = 10 red light runners are caught per day.
Let x = number of drivers caught during rush hour on a given weekday.
a. i. P(x = 14) =
 x e 
x!

(10 )14 e 10 1000000000 000000 .0000453999

14!
87 ,178 ,291,200
ii. Using Table VI of Appendix C of the text, we obtain:
  .0521
100
Chapter Five
P(at most 7) = P(0) + P(1) + P(2) + P(3) + P(4) +P(5) + P(6) + P(7) = .0000 + .0005 + .0023 +
.0076 + .0189 + .0378 + .0631 + .0901 = .2203
iii. P(13 to 18) = P(13) + P(14) + P(15) + P(16) + P(17) +P(18) = .0729 + .0521 + .0347 + .0217 +
.0128 + .0071= .2013
b.
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
P(x)
.0000
.0005
.0023
.0076
.0189
.0378
.0631
.0901
.1126
.1251
.1251
.1137
.0948
.0729
.0521
.0347
.0217
.0128
.0071
.0037
.0019
.0009
.0004
.0002
.0001
19. See solution to Exercise 5.57.