Chapter 5
Discrete Probability Distributions
Learning Objectives
1.
Understand the concepts of a random variable and a probability distribution.
2.
Be able to distinguish between discrete and continuous random variables.
3.
Be able to compute and interpret the expected value, variance, and standard deviation for a discrete
random variable.
4.
Be able to construct an empirical discrete distribution from available data.
5.
Be able to compute the covariance and correlation coefficient for a bivariate empirical discrete
distribution.
6.
Be able to compute and work with probabilities involving a binomial probability distribution.
7.
Be able to compute and work with probabilities involving a Poisson probability distribution.
8.
Know when and how to use the hypergeometric probability distribution.
5-1
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Chapter 5
Solutions:
1.
a.
Head, Head (H,H)
Head, Tail (H,T)
Tail, Head (T,H)
Tail, Tail (T,T)
b.
x = number of heads on two coin tosses
c.
Outcome
(H,H)
(H,T)
(T,H)
(T,T)
2.
Values of x
2
1
1
0
d.
Discrete. It may assume 3 values: 0, 1, and 2.
a.
Let x = time (in minutes) to assemble the product.
b.
It may assume any positive value: x > 0.
c.
Continuous
3.
Let Y = position is offered
N = position is not offered
a.
S = {(Y,Y,Y), (Y,Y,N), (Y,N,Y), (N,Y,Y), (Y,N,N), (N,Y,N), (N,N,Y), (N,N,N)}
b.
Let N = number of offers made; N is a discrete random variable.
c.
Experimental Outcome
Value of N
4.
5.
(Y,Y,Y) (Y,Y,N) (Y,N,Y) (N,Y,Y) (Y,N,N) (N,Y,N) (N,N,Y) (N,N,N)
3
2
2
2
1
1
1
0
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
a.
S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
b.
Experimental Outcome
Number of Steps Required
6.
a.
values: 0,1,2,...,20
discrete
b.
values: 0,1,2,...
discrete
c.
values: 0,1,2,...,50
discrete
values: 0  x  8
continuous
d.
(1,1)
2
(1,2)
3
(1,3)
4
(2,1)
3
(2,2)
4
(2,3)
5
5-2
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Discrete Probability Distributions
7.
e.
values: x > 0
continuous
a.
f (x)  0 for all values of x.
 f (x) = 1 Therefore, it is a proper probability distribution.
8.
b.
Probability x = 30 is f (30) = .25
c.
Probability x  25 is f (20) + f (25) = .20 + .15 = .35
d.
Probability x > 30 is f (35) = .40
a.
x
1
2
3
4
f (x)
3/20 = .15
5/20 = .25
8/20 = .40
4/20 = .20
Total 1.00
b.
c.
f (x)  0 for x = 1,2,3,4.
 f (x) = 1
5-3
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Chapter 5
9.
a.
There are a total of 26,975 unemployed persons in the data set. Each probability f(x) is computed by
dividing the number of months of unemployment by 26,975. For example, f (1) = 1029/26,975 =
.0381. The complete probability distribution is as follows.
x
1
2
3
4
5
6
7
8
9
10
b.
c.
f (x)
.0381
.0625
.0841
.0992
.1293
.1725
.1537
.1330
.0862
.0415
f ( x)  0 and  f ( x)  1
Probability 2 months or less = f (1) + f (2) = .0381 + .0625 = .1006
Probability more than 2 months = 1 - .1006 = .8994
d.
Probability more than 6 months = f (7) + f (8) + f (9) + f (10) = .1537 + .1330 + .0862 + .0415 =
.4144
10. a.
x
1
2
3
4
5
f (x)
0.05
0.09
0.03
0.42
0.41
1.00
x
1
2
3
4
5
f (x)
0.04
0.10
0.12
0.46
0.28
1.00
b.
c.
P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83
d.
Probability of very satisfied: 0.28
e.
Senior executives appear to be more satisfied than middle managers. 83% of senior executives have
a score of 4 or 5 with 41% reporting a 5. Only 28% of middle managers report being very satisfied.
5-4
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Discrete Probability Distributions
11. a.
Duration of Call
x
1
2
3
4
f (x)
0.25
0.25
0.25
0.25
1.00
b.
f (x)
0.30
0.20
0.10
x
0
1
2
3
4
c.
f (x)  0 and f (1) + f (2) + f (3) + f (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00
d.
f (3) = 0.25
e.
P(overtime) = f (3) + f (4) = 0.25 + 0.25 = 0.50
12. a.
Yes; f (x)  0.  f (x) = 1
b.
f (500,000) + f (600,000) = .10 + .05 = .15
c.
f (100,000) = .10
13. a.
Yes, since f (x)  0 for x = 1,2,3 and  f (x) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1
b.
f (2) = 2/6 = .333
c.
f (2) + f (3) = 2/6 + 3/6 = .833
14. a.
f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150)
= 1 - .95 = .05
This is the probability MRA will have a $200,000 profit.
b.
P(Profit) = f (50) + f (100) + f (150) + f (200)
= .30 + .25 + .10 + .05 = .70
c.
P(at least 100) = f (100) + f (150) + f (200)
= .25 + .10 +.05 = .40
5-5
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Chapter 5
15. a.
x
3
6
9
f (x)
.25
.50
.25
1.00
x f (x)
.75
3.00
2.25
6.00
E(x) =  = 6
b.
x-
-3
0
3
x
3
6
9
(x - )2
9
0
9
f (x)
.25
.50
.25
(x - )2 f (x)
2.25
0.00
2.25
4.50
Var(x) = 2 = 4.5
c.
 =
4.50 = 2.12
16. a.
y
2
4
7
8
f (y)
.2
.3
.4
.1
1.0
y f (y)
.4
1.2
2.8
.8
5.2
E(y) =  = 5.2
b.
y-
-3.20
-1.20
1.80
2.80
y
2
4
7
8
(y - )2
10.24
1.44
3.24
7.84
(y - )2 f (y)
2.048
.432
1.296
.784
4.560
f (y)
.20
.30
.40
.10
Var ( y )  4.56
  4.56  2.14
17. a.
Total Student = 1,518,859
x=1
x=2
x=3
x=4
x=5
b.
f(1) = 721,769/1,518,859 = .4752
f(2) = 601,325/1,518,859 = .3959
f(3) = 166,736/1,518,859 = .1098
f(4) = 22,299/1,518,859 = .0147
f(5) =
6730/1,518,859 = .0044
P(x > 1) = 1 – f(1) = 1 - .4752 = .5248
Over 50% of the students take the SAT more than 1 time.
5-6
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Discrete Probability Distributions
c.
P(x > 3) = f(3) + f(4) + f(5) = .1098 + .0147 + .0044 = .1289
d./e.
x
1
2
3
4
5
f (x)
.4752
.3959
.1098
.0147
.0044
x f (x)
.4752
.7918
.3293
.0587
.0222
1.6772
x-
-.6772
.3228
1.3228
2.3228
3.3228
(x - )2
.4586
.1042
1.7497
5.3953
11.0408
(x - )2 f (x)
.2179
.0412
.1921
.0792
.0489
.5794
E(x) = Σ x f(x) = 1.6772
The mean number of times a student takes the SAT is 1.6772, or approximately
1.7 times.
 2  ( x   ) 2 f ( x)  .5794
   2  .5794  .7612
18. a/b.
x
0
1
2
3
4
Total
f (x)
0.04
0.34
0.41
0.18
0.04
1.00
xf (x)
0.00
0.34
0.82
0.53
0.15
1.84

E(x)
x - 
-1.84
-0.84
0.16
1.16
2.16
(x - )2
3.39
0.71
0.02
1.34
4.66
(x - )2 f (x)
0.12
0.24
0.01
0.24
0.17
0.79

Var(x)
y
0
1
2
3
4
Total
f (y)
0.00
0.03
0.23
0.52
0.22
1.00
yf (y)
0.00
0.03
0.45
1.55
0.90
2.93

E(y)
y - 
-2.93
-1.93
-0.93
0.07
1.07
(y - )2
8.58
3.72
0.86
0.01
1.15
(y - )2 f (y)
0.01
0.12
0.20
0.00
0.26
0.59

Var(y)
c/d.
e.
19. a.
The number of bedrooms in owner-occupied houses is greater than in renter-occupied houses. The
expected number of bedrooms is 2.93 - 1.84 = 1.09 greater. And, the variability in the number of
bedrooms is less for the owner-occupied houses.
E(x) =  x f (x) = 0 (.56) + 2 (.44) = .88
b.
E(x) =  x f (x) = 0 (.66) + 3 (.34) = 1.02
c.
The expected value of a 3 - point shot is higher. So, if these probabilities hold up, the team will
make more points in the long run with the 3 - point shot.
5-7
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Chapter 5
20. a.
x
0
500
1000
3000
5000
8000
10000
Total
f (x)
.85
.04
.04
.03
.02
.01
.01
1.00
xf (x)
0
20
40
90
100
80
100
430
The expected value of the insurance claim is $430. If the company charges $430 for this type of
collision coverage, it would break even.
b.
From the point of view of the policyholder, the expected gain is as follows:
Expected Gain = Expected claim payout – Cost of insurance coverage
= $430 - $520 = -$90
The policyholder is concerned that an accident will result in a big repair bill if there is no insurance
coverage. So even though the policyholder has an expected annual loss of $90, the insurance is
protecting against a large loss.
21. a.
E(x) =  x f (x) = 0.05(1) + 0.09(2) + 0.03(3) + 0.42(4) + 0.41(5) = 4.05
b.
E(x) =  x f (x) = 0.04(1) + 0.10(2) + 0.12(3) + 0.46(4) + 0.28(5) = 3.84
c.
Executives:
2 =  (x - )2 f(x) = 1.25
Middle Managers: 2 =  (x - )2 f(x) = 1.13
d.
Executives:
 = 1.12
Middle Managers:  = 1.07
e.
22. a.
The senior executives have a higher average score: 4.05 vs. 3.84 for the middle managers. The
executives also have a slightly higher standard deviation.
E(x) =  x f (x) = 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445
The monthly order quantity should be 445 units.
b.
23. a.
Cost:
445 @ $50 = $22,250
Revenue: 300 @ $70 = 21,000
$ 1,250 Loss
Rent Controlled: E(x) = 1(.61) + 2(.27) + 3(.07) + 4(.04) + 5(.01) = 1.57
Rent Stabilized: E(x) = 1(.41) + 2(.30) + 3(.14) + 4(.11) + 5(.03) + 6(.01) = 2.08
b.
Rent Controlled:
Var(x) = (-.57)2.61+ (.43)2.27+ (1.43)2.07+ (2.43)2.04+ (3.43)2.01= .75
5-8
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Discrete Probability Distributions
Rent Stabilized:
Var(x) = (-1.08)2.41+ (-.08)2.30+ (.92)2.14+ (1.92)2.11+ (2.92)2.03+ (3.92)2.01= 1.41
c.
24. a.
From the expected values in part (a), it is clear that the expected number of persons living in rent
stabilized units is greater than the number of persons living in rent controlled units. For example,
comparing a building that contained 10 rent controlled units to a building that contained 10 rent
stabilized units, the expected number of persons living in the rent controlled building would be
1.57(10) = 15.7 or approximately 16. For the rent stabilized building, the expected number of
persons is approximately 21. There is also more variability in the number of persons living in rent
stabilized units.
Medium E(x) =  x f (x)
= 50 (.20) + 150 (.50) + 200 (.30) = 145
E(x) =  x f (x)
Large:
= 0 (.20) + 100 (.50) + 300 (.30) = 140
Medium preferred.
b.


Medium

x
50
150
200
 
  
f (x)
.20
.50
.30
x-
-95
5
55
Large
y
0
100
300
(x - )2
9025
25
3025

y-
-140
-40
160
f (y)
.20
.50
.30
(y - )2
19600
1600
25600
(x - )2 f (x)
1805.0
12.5
907.5
2 = 2725.0
(y - )2 f (y)
3920
800
7680
2 = 12,400
Medium preferred due to less variance.
25. a.
E ( x)  .2(50)  .5(30)  .3(40)  37
E ( y)  .2(80)  .5(50)  .3(60)  59
Var ( x)  .2(50  37) 2  .5(30  37) 2  .3(40  37) 2  61
Var ( y )  .2(80  59) 2  .5(50  59) 2  .3(60  59) 2  129
b.
x+y
130
80
100
f (x +y)
.2
.5
.3
5-9
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Chapter 5
c.
x+y
f(x +y)
(x + y)f(x + y)
x + y – E(x + y)
130
80
100
.2
.5
.3
E(x + y) =
26
40
30
96
34
-16
4
d.
[ x  y  E ( x  y )]2
1156
256
16
Var(x + y) =
[ x  y  E ( x  y )]2 f ( x  y )
231.2
128.0
4.8
364
 xy  [Var ( x  y )  Var ( x)  Var ( y)] / 2  (364  61  129) / 2  87
Var(x) = 61 and Var(y) =129 were computed in part (a), so
 x  61  7.8102
 xy 
 y  129  11.3578
 xy
87

 .98
 x y (7.8102)(11.3578)
The random variables x and y are positively related. Both the covariance and correlation coefficient
are positive. Indeed, they are very highly correlated; the correlation coefficient is almost equal to 1.
e.
Var ( x  y )  Var ( x)  Var ( y )  2 xy  61  129  2(87)  364
Var(x) + Var(y) = 61 + 129 = 190
The variance of the sum of x and y is greater than the sum of the variances by two times the
covariance: 2(87) = 174. The reason it is positive is that, in this case the variables are positively
related. Whenever two random variables are positively related, the variance of the sum of the
randomly variables will be greater than the sum of the variances of the individual random variables.
26. a.
The standard deviation for these two stocks is the square root of the variance.
 x  Var ( x)  25  5%
 y  Var ( y)  1  1%
Investments in Stock 1 would be considered riskier than investments in Stock 2 because the standard
deviation is higher. Note that if the return for Stock 1 falls 8.45/5 = 1.69 or more standard deviation
below its expected value, an investor in that stock will experience a loss. The return for Stock 2
would have to fall 3.2 standard deviations below its expected value before an investor in that stock
would experience a loss.
b.
Since x represents the percent return for investing in Stock 1, the expected return for investing $100
in Stock 1 is $8.45 and the standard deviation is $5.00. So to get the expected return and standard
deviation for a $500 investment we just multiply by 5.
Expected return ($500 investment) = 5($8.45) = $42.25
Standard deviation ($500 investment) = 5($5.00) = $25.00
c.
Since x represents the percent return for investing in Stock 1 and y represents the percent return for
investing in Stock 2, we want to compute the expected value and variance for .5x + .5y.
5 - 10
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Discrete Probability Distributions
E(.5x + .5y) = .5E(x) + .5E(y) = .5(8.45) + .5(3.2) = 4.225 + 1.6 = 5.825
Var (.5 x  .5 y )  .52Var ( x)  .52Var ( y )  2(.5)(.5) xy
 (.5) 2 (25)  (.5) 2 (1)  2(.5)(.5)( 3)
 6.25  .25  1.50  5
 .5 x.5 y  5  2.236
d.
Since x represents the percent return for investing in Stock 1 and y represents the percent return for
investing in Stock 2, we want to compute the expected value and variance for .7x + .3y.
E(.7x + .3y) = .7E(x) + .3E(y) = .7(8.45) + .3(3.2) =5.915 + .96 = 6.875
Var (.7 x  .3 y )  .7 2Var ( x)  .32Var ( y )  2(.7)(.3) xy
 .7 2 (25)  .32 (1)  2(.7)(.3)(3)
 12.25  .09  1.26  11.08
 .7 x.3 y  11.08  3.329
e.
The standard deviations of x and y were computed in part (a). The correlation coefficient is given by
 xy 
 xy
3

 .6
 x y (5)(1)
There is a fairly strong negative relationship between the variables.
27. a.
Dividing each of the frequencies in the table by the total number of restaurants provides the joint
probability table below. The bivariate probability for each pair of quality and meal price is shown in
the body of the table. This is the bivariate probability distribution. For instance, the probability of a
rating of 2 on quality and a rating of 3 on meal price is given by f(2, 3) = .18. The marginal
probability distribution for quality, x, is in the rightmost column. The marginal probability for meal
price, y, is in the bottom row.
Meal Price (y)
b.
Quality (x)
1
2
3
Total
1
0.14
0.13
0.01
0.28
2
0.11
0.21
0.18
0.50
3
0.01
0.05
0.16
0.22
Total
0.26
0.39
0.35
1
E(x) = 1(.28) + 2(.50) + 3(.22) = 1.94
Var(x) = .28(1 – 1.94)2 + .50(2- 1.94)2 + .22(3 – 1.94)2 = .4964
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Chapter 5
c.
E(y) = 1(.26) + 2(.39) + 3(.35) = 2.09
Var(y) = .26(1 – 2.09)2 + .39(2 – 2.09)2 + .35(3 – 2.09)2 = .6019
d.
 xy  [Var( x  y )  Var( x)  Var( y)] / 2  [1.6691  .4964  .6019] / 2  .2854
Since, the covariance  xy  .2854 is positive we can conclude that as the quality rating goes up, the
meal price goes up. This is as we would expect.
e.
 xy 
 xy
.2854

 .5221
 x y
.4964 .6019
With a correlation coefficient of .5221 we would call this a moderately positive relationship. It is
not likely to find a low cost restaurant that is also high quality. But, it is possible. There are 3 of
them leading to f (3,1)  .01.
28. a.
Marginal distribution of Direct Labor Cost
y
f (y)
yf (y)
y - E(y)
(y-E(y))2
(y-E(y))2f (y)
43
.3
12.9
-2.3
5.29
1.587
45
.4
18
-.3
.09
.036
48
.3
14.4
2.7
7.29
2.187
45.3
Var(y)=
3.81
y=
1.95
E(y) = 45.3
b.
Marginal distribution of Parts Cost
x
f (x)
xf (x)
x - E(x)
(x-E(x))2
85
.45
38.25
-5.5
30.25
13.6125
95
.55
52.25
4.5
20.25
11.1375
90.5
E(x) = 90.5
c.
(x-E(x))2f (x)
Var(x)=
24.75
x =
4.97
Let z = x + y represent total manufacturing cost (direct labor + parts).
z
128
130
133
138
140
143
f (z)
.05
.20
.20
.25
.20
.10
1.00
5 - 12
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Discrete Probability Distributions
d.
The computation of the expected value, variance, and standard deviation of total manufacturing cost
is shown below.
z - E(z)
(z-E(z))2
(z-E(z))2f(z)
6.4
-7.8
60.84
3.042
.20
26
-5.8
33.64
6.728
133
.20
26.6
-2.8
7.84
1.568
138
.25
34.5
2.2
4.84
1.21
140
.20
28
4.2
17.64
3.528
143
.10
14.3
7.2
51.84
Var(z)=
5.184
z
f (z)
zf (z)
128
.05
130
135.8
E(z) = 135.8
e.
z =
21.26
4.61
To determine if x = parts cost and y = direct labor cost are independent, we need to compute the
covariance  xy .
 xy  (Var( x  y)  Var ( x) - Var ( y)) / 2  (21.26 - 24.75 - 3.81) / 2  -3.65
Since the covariance is not equal to zero, we can conclude that direct labor cost is not independent of
parts cost. Indeed, they are negatively correlated. When parts cost goes up, direct labor cost goes
down. Maybe the parts costing $95 come from a different manufacturer and are higher quality.
Working with higher quality parts may reduce labor costs.
f.
The expected manufacturing cost for 1500 printers is
E (1500 z)  1500E( z)  1500(135.8)  203,700
The total manufacturing costs of $198,350 are less than we would have expected. Perhaps as more
printers were manufactured there was a learning curve and direct labor costs went down.
29. a.
Let x = percentage return for S&P 500
y = percentage return for Core Bond fund
The formula for computing the correlation coefficient is given by
 xy 
 xy
 x y
In this case, we know the correlation coefficient and both standard deviations, so we want to
rearrange this formula to find the covariance.
 xy   xy x y  (.32)(19.45)(2.13)  13.26
b.
Letting r = portfolio percentage return, we have r = .5x + .5y. The expected return for a portfolio
with 50% invested in the S&P 500 and 50% invested in Core Bonds is
E (r )  .5E ( x)  .5E ( y)  (.5)5.04%  (.5)5.78%  5.41%
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Chapter 5
We are given  x and  y , so Var ( x)  19.452  378.3025 and Var ( y )  2.132  4.5369 . We can now
compute
Var (.5 x  .5 y )  .52Var ( x)  .52Var ( y )  2(.5)(.5)( xy )
 .25(378.3025)  .25(4.5369)  .5( 13.26)
 89.08
Then  x y  89.08  9.44
So, the expected return for our portfolio is 5.41% and the standard deviation is 9.78%.
c.
Letting r = portfolio percentage return, we have r = .2x + .8y. The expected return for a portfolio
with 20% invested in the S&P 500 and 80% invested in Core Bonds is
E (r )  .2E( x)  .8E( y)  (.2)5.04%  (.8)5.78%  5.63%
We are given  x and  y , so Var ( x)  19.452  378.3025 and Var ( y )  2.132  4.5369 . We can now
compute
Var (.2 x  .8 y )  .22Var ( x)  .82Var ( y )  2(.2)(.8)( xy )
 .04(378.3025)  .64(4.5369)  .32( 13.26)
 13.79
Then  x y  13.79  3.71
So, the expected return for our portfolio is 5.63% and the standard deviation is 3.71%.
d.
Letting r = portfolio percentage return, we have r = .8x + .2y. The expected return for a portfolio
with 80% invested in the S&P 500 and 20% invested in Core Bonds is
E (r )  .8E( x)  .2E( y)  (.8)5.04%  (.2)5.78%  5.19%
We are given  x and  y , so Var ( x)  19.452  378.3025 and Var ( y )  2.132  4.5369 .
We can now compute
Var (.8 x  .2 y )  .82Var ( x)  .22Var ( y )  2(.8)(.2)( xy )
 .64(378.3025)  .04(4.5369)  .32( 13.26)
 238.05
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Discrete Probability Distributions
Then  x y  238.05  15.43
So, the expected return for our portfolio is 5.19% and the standard deviation is 15.43%.
e.
The portfolio in part (c), investing 20% in the S&P500 fund and 80% in the Core Bond fund has the
largest expected return: 5.63%.
The portfolio in part (c), investing 20% in the S&P 500 fund and 80% in the Core Bond fund also
has the smallest standard deviation: 3.71%.
Since the portfolio in part (c) has the highest expected return and the smallest standard deviation (our
measure of risk) it is the preferred investment choice.
30. a.
Let x = percentage return for S&P 500
y = percentage return for Core Bond fund
z = percentage return for REITs
The formula for computing the correlation coefficient is given by
 xy 
 xy
 x y
In this case, we know the correlation coefficients and the 3 standard deviations, so we want to
rearrange the correlation coefficient formula to find the covariances.
S&P 500 and REITs:  xz   xz x z  (.74)(19.45)(23.17)  333.486
Core Bonds and REITS:  yz   yz y z  (.04)(2.13)(23.17)  1.974
b.
Letting r = portfolio percentage return, we have r = .5x + .5y. The expected return for a portfolio
with 50% invested in the S&P 500 and 50% invested in REITs is
E (r )  .5E ( x)  .5E ( z )  (.5)5.04%  (.5)13.07%  9.055%
We are given  x and  y , so Var ( x)  19.452  378.3025 and Var ( z )  23.17 2  536.8489 . We can
now compute
Var (.5 x  .5 z )  .52Var ( x)  .52Var ( z )  2(.5)(.5)( xz )
=.25(378.3025)+.25(536.8489)+.5(333.486)
=395.53
Then  x y  395.53  19.89
So, the expected return for our portfolio is 9.055% and the standard deviation is 19.89%.
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Chapter 5
c.
Letting r = portfolio percentage return, we have r = .5y + .5z. The expected return for a portfolio
with 50% invested in Core Bonds and 50% invested in REITs is
E (r )  .5E ( y)  .5E ( z )  (.5)5.78%  (.5)13.07%  9.425%
We are given  x and  y , so Var ( y )  2.132  4.5369 and Var ( z )  23.17 2  536.8489 . We can now
compute
Var (.5 y  .5 z )  .52Var ( y )  .52Var ( z )  2(.5)(.5)( yz )
 .25(4.5369)  .25(536.8489)  .5(.04)
 135.33
Then  y  z  135.33  11.63
So, the expected return for our portfolio is 9.425% and the standard deviation is 11.63%.
d.
Letting r = portfolio percentage return, we have r = .8y + .2z. The expected return for a portfolio
with 80% invested in Core Bonds and 20% invested in REITs is
E (r )  .8E( y)  .2E( z)  (.8)5.78%  (.2)13.07%  7.238%
From part (c) above, we have Var ( y)  4.5369 and Var ( z )  536.8489 . We can now compute
Var (.8 y  .2 z )  .82 Var ( y )  .22 Var ( z )  2(.8)(.2)( yz )
 .64(4.5369)  .04(536.8489)  .32(.04)
 24.36
Then  y  z  24.36  4.94
So, the expected return for our portfolio is 7.238% and the standard deviation is 4.94%.
e.
The expected returns and standard deviations for the 3 portfolios are summarized below.
Portfolio
50% S&P 500 & 50% REITs
50% Core Bonds & 50% REITs
80% Core Bonds & 20% REITs
Expected Return (%)
9.055
9.425
7.238
Standard Deviation
19.89
11.63
4.94
The portfolio from part (c) involving 50% Core Bonds and 50% REITS has the highest return.
Using the standard deviation as a measure of risk, it also has less risk than the portfolio from part (b)
involving 50% invested in an S&P 500 index fund and 50% invested in REITs. So the portfolio
from part (b) would not be recommended for either type of investor.
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Discrete Probability Distributions
The portfolio from part (d) involving 80% in Core Bonds and 20% in REITs has the lowest standard
deviation and thus lesser risk than the portfolio in part (c). We would recommend the portfolio
consisting of 50% Core Bonds and 50% REITs for the aggressive investor because of its higher
return and moderate amount of risk.
We would recommend the portfolio consisting of 80% Core Bonds and 20% REITS to the
conservative investor because of its low risk and moderate return.
31. a.
b.
 2
2!
f (1)    (.4)1 (.6)1 
(.4)(.6)  .48
1
1!1!
 
c.
 2
2!
f (0)    (.4)0 (.6) 2 
(1)(.36)  .36
0!2!
0
d.
 2
2!
f (2)    (.4) 2 (.6)0 
(.16)(1)  .16
2
2!0!
 
e.
P(x  1) = f (1) + f (2) = .48 + .16 = .64
f.
E(x) = n p = 2 (.4) = .8
Var(x) = n p (1 - p) = 2 (.4) (.6) = .48
 =
.48 = .6928
32. a.
f (0) = .3487
b.
f (2) = .1937
c.
P(x  2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298
d.
P(x  1) = 1 - f (0) = 1 - .3487 = .6513
e.
E(x) = n p = 10 (.1) = 1
f.
Var(x) = n p (1 - p) = 10 (.1) (.9) = .9
 = .9 = .95
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Chapter 5
33. a.
f (12) = .1144
b.
f (16) = .1304
c.
P(x  16) = f (16) + f (17) + f (18) + f (19) + f (20)
= .1304 + .0716 + .0278 + .0068 + .0008 = .2374
d.
P(x  15) = 1 - P (x  16) = 1 - .2374 = .7626
e.
E(x) = n p = 20(.7) = 14
f.
Var(x) = n p (1 - p) = 20 (.7) (.3) = 4.2
 =
34. a.
b.
4.2 = 2.0494
6
f (2)    (.23) 2 (.77) 4  .2789
 2
P(at least 2) = 1 - f(0) - f(1)
6
 6
= 1    (.23)0 (.77)6    (.23)1 (.77)5
0
1
=
c.
35. a.
b.
1 - .2084 - .3735 = .4181
10 
f (0)    (.23)0 (.77)10  .0733
0
n
f ( x)    ( p) x (1  p) n  x
 x
f (3) 
10!
(.30)3 (1  .30)10 3
3!(10  3)!
f (3) 
10(9)(8)
(.30)3 (1  .30)7  .2668
3(2)(1)
P(x  3) = 1 - f (0) - f (1) - f (2)
f (0) 
10!
(.30)0 (1  .30)10  .0282
0!(10)!
f (1) 
10!
(.30)1 (1  .30)9  .1211
1!(9)!
f (2) 
10!
(.30) 2 (1  .30)8  .2335
2!(8)!
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Discrete Probability Distributions
P(x  3) = 1 - .0282 - .1211 - .2335 = .6172
36. a.
b.
Probability of a defective part being produced must be .03 for each part selected; parts must be
selected independently.
Let: D = defective
G = not defective
1st part
Experimental
Outcome
2nd part
D
Number
Defective
(D, D)
2
(D, G)
1
(G, D)
1
(G, G)
0
G
D
.
G
D
G
c.
2 outcomes result in exactly one defect.
d.
P(no defects) = (.97) (.97) = .9409
P (1 defect) = 2 (.03) (.97) = .0582
P (2 defects) = (.03) (.03) = .0009
37. a.
Yes. Since the employees are selected randomly, p is the same from trial to trial and the trials are
independent. The two outcomes per trial are loyal and not loyal.
Binomial n = 10 and p = .25
f ( x) 
10!
(.25) x (1  .25)10  x
x !(10  x)!
b.
f (0) 
10!
(.25)0 (1  .25)10  0  .0563
0!(10  0)!
c.
f (4) 
10!
(.25) 4 (1  .25)10  4  .1460
4!(10  4)!
d.
Probability (x > 2) = 1 - f (0) - f (1)
From part (b), f(0) = .0563
f (1) 
10!
(.25)1 (1  .25)10 1  .1877
1!(10  1)!
Probability (x > 2) = 1 - f (0) - f (1) = 1 - (.0563 + .1877 ) = .7560
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Chapter 5
38. a.
b.
.90
P(at least 1) = f (1) + f (2)
f (1) 
2!
(.9)1 (.1)1
1! 1!
 2(.9)(.1)  .18
f (2) 
2!
(.9)1 (.1)0
2! 0!
 1(.81)(1)  .81
 P(at least 1) = .18 + .81 = .99
Alternatively
P(at least 1) = 1 – f (0)
f (0) 
2!
(.9) 0 (.1) 2  .01
0! 2!
Therefore, P(at least 1) = 1 - .01 = .99
c.
P(at least 1) = 1 - f (0)
f (0) 
3!
(.9)0 (.1)3  .001
0! 3!
Therefore, P(at least 1) = 1 - .001 = .999
d.
39. a.
Yes; P(at least 1) becomes very close to 1 with multiple systems and the inability to detect an attack
would be catastrophic.
Using the 20 golfers in the Hazeltine PGA Championship, the probability that a PGA professional
golfer uses a Titleist brand golf ball is p = 14/20 = .6
For the sample of 15 PGA Tour players, use a binomial distribution with n = 15 and p = .6.
f (10) =
15!
(.6)10 (1  .6)15 10 =.1859
10!5!
Or, using the binomial tables, f (10) = .1859
b.
P(x > 10) = f (11) + f (12) + f (13) + f (14) + f (15)
Using the binomial tables, we have
.1268 + .0634 + .0219 + .0047 + .0005 = .2173
c.
E(x) = np = 15(.6) = 9
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Discrete Probability Distributions
d.
Var(x) = 2 = np(1 - p) = 15(.6)(1 - .6) = 3.6
 =
40. a.
b.
3.6 = 1.8974
n
f ( x)    ( p) x (1  p) n  x
 x
f (4) 
15!
(.28) 4 (1  .28)15 4
4!(15  4)!
f (4) 
15(14)(13)(12)
(.28)4 (1  .28)11  .2262
4(3)(2)(1)
P(x  3) = 1 - f (0) - f (1) - f (2)
f (0) 
15!
(.28)0 (1  .28)15  .0072
0!(15)!
f (1) 
15!
(.28)1 (1  .28)14  .0423
1!(14)!
f (2) 
15!
(.28) 2 (1  .28)13  .1150
2!(13)!
P(x  3) = 1 - .0072 - .0423 - .1150 = .8355
41. a.
f (0) + f (1) + f (2) = .0115 + .0576 + .1369 = .2060
b.
f (4) = .2182
c.
1 - [ f (0) + f (1) + f (2) + f (3) ]
d.
 = n p = 20 (.20) = 4
42. a.
= 1 - .2060 - .2054 = .5886
p = ¼ = .25
n
f ( x)    ( p) x (1  p) n  x
 x
 20 
f (4)    (.25) 4 (1  .25) 20  4
4
f (4) 
b.
20!
20(19)(18)(17)
(.25) 4 (.75)16  f (4) 
(.25) 4 (.75)16  .1897
4!(20  4)!
4(3)(2)(1)
P(x  2) = 1 – f(0) – f(1)
Using the binomial tables f(0) = .0032 and f(1) = .0211
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Chapter 5
P(x  2) = 1 – .0032 - .0211 = .9757
c.
Using the binomial tables f(12) = .0008
And, with f (13) = .0002, f (14) = .0000, and so on, the probability of finding that 12 or more
investors have exchange-traded funds in their portfolio is so small that it is highly unlikely that p =
.25. In such a case, we would doubt the accuracy of the results and conclude that p must be greater
than .25.
d.
43.
 = n p = 20 (.25) = 5
E(x) = n p = 35(.23) = 8.05 (8 automobiles)
Var(x) = n p (1 - p) = 35(.23)(1-.23) = 6.2
 =
6.2 = 2.49
44. a.
f ( x) 
3x e3
x!
b.
f (2) 
32 e3 9(.0498)

 .2241
2!
2
c.
f (1) 
31 e3
 3(.0498)  .1494
1!
d.
45. a.
b.
P(x  2) = 1 - f (0) - f (1) = 1 - .0498 - .1494 = .8008
f ( x) 
2 x e2
x!
 = 6 for 3 time periods
c.
f ( x) 
6 x e6
x!
d.
f (2) 
22 e2 4(.1353)

 .2706
2!
2
e.
f (6) 
66 e6
 .1606
6!
f.
f (5) 
45 e4
 .1563
5!
46. a.
 = 48 (5/60) = 4
3
-4
f (3) = 4 e
3!
b.
= (64) (.0183) = .1952
6
 = 48 (15 / 60) = 12
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Discrete Probability Distributions
10
-12
f (10) = 12 e
10 !
c.
= .1048
 = 48 (5 / 60) = 4 I expect 4 callers to be waiting after 5 minutes.
0
-4
f (0) = 4 e
0!
= .0183
The probability none will be waiting after 5 minutes is .0183.
d.
 = 48 (3 / 60) = 2.4
0
f (0) = 2.4 e
0!
-2.4
= .0907
The probability of no interruptions in 3 minutes is .0907.
47. a.
b.
30 per hour
 = 1 (5/2) = 5/2
f (3) 
(5 / 2)3 e(5 / 2)
 .2138
3!
c.
f (0) 
(5 / 2)0 e(5 / 2)
 e(5 / 2)  .0821
0!
48. a.
f (0) 
70 e7
 e7  .0009
0!
b.
probability = 1 - [f(0) + f(1)]
f (1) 
71 e7
 7e7  .0064
1!
probability = 1 - [.0009 + .0064] = .9927
c.
 = 3.5
f (0) 
3.50 e3.5
 e3.5  .0302
0!
probability = 1 - f(0) = 1 - .0302 = .9698
d.
probability = 1 - [f(0) + f(1) + f(2) + f(3) + f(4)]
= 1 - [.0009 + .0064 + .0223 + .0521 + .0912] = .8271
Note: The Poisson tables were used to compute the Poisson probabilities f(0), f(1), f(2), f(3) and f(4)
in part (d).
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Chapter 5
49. a.
b.
f (0) 
100 e10
 e10  .000045
0!
f (0) + f (1) + f (2) + f (3)
f (0) = .000045 (part a)
f (1) 
101 e10
 .00045
1!
Similarly, f (2) = .002267, f (3) = .007567
and f (0) + f (1) + f (2) + f (3) = .010329
c.
2.5 arrivals / 15 sec. period Use  = 2.5
f (0) 
d.
50.
2.50 e2.5
 .0821
0!
1 - f (0) = 1 - .0821 = .9179
Poisson distribution applies
a.
b.
c.
d.
51. a.
 = 1.25 per month
1.250 e1.25
 .2865
0!
1.251 e1.25
f (1) 
 .3581
1!
f (0) 
P(More than 1) = 1 - f (0) - f (1) = 1 - 0.2865 - 0.3581 = .3554
f ( x) 
f (0) 
b.
 x e 
x!
30 e3
 e3  .0498
0!
P(x  2) = 1 - f (0) - f (1)
f (1) 
31 e3
 .1494
1!
P(x  2) = 1 - .0498 - .1494 = .8008
c.
µ = 3 per year
µ = 3/2 = 1.5 per 6 months
d.
f (0) 
1.50 e1.5
 .2231
0!
5 - 24
© 2013 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Discrete Probability Distributions
52. a.
 3  10  3 
 

1 4 1 
f (1)   

10 
 
4
 3!  7! 



 1!2!  3!4!   (3)(35)  .50
10!
210
4!6!
b.
 3  10  3 
 

2 2  2  (3)(1)
f (2)   

 .067
45
 10 
 
2
c.
 3 10  3 
 

0 2  0  (1)(21)
f (0)   

 .4667
45
10 
 
2
d.
 3 10  3 
 

2 4  2  (3)(21)
f (2)   

 .30
210
10 
 
4
e.
Note x = 4 is greater than r = 3. It is not possible to have x = 4 successes when there are only 3
successes in the population. Thus, f(4) = 0. In this exercise, n is greater than r. Thus, the number of
successes x can only take on values up to and including r = 3. Thus, x = 0, 1, 2, 3.
 4   15  4 
 

3 10  3  (4)(330)
f (3)    

 .4396
3003
 15 
 
 10 
53.
54.
Hypergeometric Distribution with N = 10 and r = 7
a.
 7   3
  
2 1
(21)(3)
f (2)      
= .5250
120
 10 
 
3
b.
Compute the probability that 3 prefer football.
 7 3
  
3 0
(35)(1)
f (3)      
 .2917
120
 10 
 
3
P(majority prefer football) = f (2) + f (3) = .5250 + .2917 = .8167
5 - 25
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Chapter 5
55.
Parts a, b & c involve the hypergeometric distribution with N = 52 and n = 2
a.
r = 20, x = 2
 20  32 
  
2
0
(190)(1)
f (2)     
 .1433
1326
 52 
 
2
b.
r = 4, x = 2
 4   48 
  
2 0
(6)(1)
f (2)      
 .0045
1326
 52 
 
2
c.
r = 16, x = 2
 16  36 
  
2
0
(120)(1)
f (2)     
 .0905
1326
 52 
 
2
d.
Part (a) provides the probability of blackjack plus the probability of 2 aces plus the probability of
two 10s. To find the probability of blackjack we subtract the probabilities in (b) and (c) from the
probability in (a).
P(blackjack) = .1433 - .0045 - .0905 = .0483
56.
N = 60 n = 10
a.
r = 20 x = 0
 20  40 
 40! 
   (1) 

0
10
10!30!   40!  10!50! 
f (0)      



60!
 60 
 10!30!  60! 
 
10!50!
 10 
=
b.
40  39  38  37  36  35  34  33  32  31
 .0112
60  59  58  57  56  55  54  53  52  51
r = 20 x = 1
 20  40 
  
1
9
 40!  10!50! 
f (1)      20 

  .0725
 60 
 9!31!  60! 
 
 10 
5 - 26
© 2013 Cengage Learning. All Rights Reserved.
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Discrete Probability Distributions
c.
1 - f (0) - f (1) = 1 - .0112 - .0725 = .9163
d.
Same as the probability one will be from Hawaii; .0725.
57. a.
 5 10 
  
0 3
(1)(120)
f (0)     
 .2637
455
15 
 
3
b.
 5  10 
  
1 2
(5)(45)
f (1)      
 .4945
455
 15 
 
3
c.
 5   10 
  
2 1
(10)(10)
f (2)      
 .2198
15
455
 
 
3
d.
 5  10 
  
3 0
(10)(1)
f (3)     
 .0220
455
15 
 
3
58.
Hypergeometric with N = 10 and r = 3.
a.
n = 3, x = 0
 3   7   3!   7! 
   
0 3
0!3!   3!4!  (1)(35)
f (0)       

 .2917
10!
120
 10 
 
3!7!
3
This is the probability there will be no banks with increased lending in the study.
b.
n = 3, x = 3
 3  7   3!  7! 
   
 0!7!  (1)(1)
3 0
3!0! 

f (3)      
 .0083
10!
120
10 
 
3!7!
3
This is the probability there all three banks with increased lending will be in the study. This has a
very low probability of happening.
5 - 27
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Chapter 5
c.
n = 3, x = 1
 3   7   3!   7! 
   
1 2
1!2!   2!5!  (3)(21)
f (1)       

 .5250
10!
120
 10 
 
3!7!
3
n = 3, x = 2
 3  7   3!  7! 
   
 1!6!  (3)(7)
2 1
2!1! 

f (2)      
 .1750
10!
120
 10 
 
3!7!
3
x
0
1
2
3
Total
f(x)
0.2917
0.5250
0.1750
0.0083
1.0000
f(1) = .5250 has the highest probability showing that there is over a .50 chance that there will be
exactly one bank that had increased lending in the study.
d.
P(x > 1) = 1  f (0)  1  .2917  .7083
There is a reasonably high probability of .7083 that there will be at least one bank that had increased
lending in the study.
e.
 r 
 3 
E ( x)  n    3    .90
N
 
 10 
 r 
r  N  n 
 3 
3  10  3 
 2  n  1  
  3  10  1  10  10  1   .49
 N  N  N  1 
 


   2  .49  .70
59. a/b/c.
x
1
2
3
4
5
Total
f (x)
0.07
0.21
0.29
0.39
0.04
1.00
xf (x)
0.07
0.42
0.87
1.56
0.20
3.12

E(x)
x - 
-2.12
-1.12
-0.12
0.88
1.88
(x - )2
4.49
1.25
0.01
0.77
3.53
(x - )2 f (x)
0.31
0.26
0.00
0.30
0.14
1.03

Var(x)
 = 1.03 = 1.01
5 - 28
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Discrete Probability Distributions
d.
The expected level of optimism is 3.12. This is a bit above neutral and indicates that investment
managers are somewhat optimistic. Their attitudes are centered between neutral and bullish with the
consensus being closer to neutral.
60. a/b.
x
1
2
3
4
5
Total
c.
f (x)
0.24
0.21
0.10
0.21
0.24
1.00
x - 
-2.00
-1.00
0.00
1.00
2.00
xf (x)
0.24
0.41
0.31
0.83
1.21
3.00

E(x)
For the bond fund categories: E(x) = 1.36
For the stock fund categories: E(x) = 4
(x - )2
4.00
1.00
0.00
1.00
4.00
(x - )2 f (x)
0.97
0.21
0.00
0.21
0.97
2.34

Var(x)
Var(x) = .23
Var(x) = 1.00
The total risk of the stock funds is much higher than for the bond funds. It makes sense to analyze
these separately. When you do the variances for both groups (stocks and bonds), they are reduced.
61. a.
x
9
10
11
12
13
b.
f (x)
.30
.20
.25
.05
.20
E(x) = x f (x)
= 9(.30) + 10(.20) + 11(.25) + 12(.05) + 13(.20) = 10.65
Expected value of expenses: $10.65 million
c.
Var(x) = (x - )2 f (x)
= (9 - 10.65)2 (.30) + (10 - 10.65)2 (.20) + (11 - 10.65)2 (.25)
+ (12 - 10.65)2 (.05) + (13 - 10.65)2 (.20) = 2.13
d.
Looks Good: E(Profit) = 12 - 10.65 = 1.35 million
However, there is a .20 probability that expenses will equal $13 million and the college will run a
deficit.
5 - 29
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Chapter 5
62. a.
There are 600 observations involving the two variables. Dividing the entries in the table shown by
600 and summing the rows and columns we obtain the following.
Reading Material (y)
Snacks (x)
0
1
2
Total
0
.0
.1
.03
.13
1
.4
.15
.05
.6
2
.2
.05
.02
.27
Total
.6
.3
.1
1
The entries in the body of the table are the bivariate or joint probabilities for x and y. The entries in
the right most (Total) column are the marginal probabilities for x and the entries in the bottom
(Total) row are the marginal probabilities for y.
The probability of a customer purchasing 1 item of reading materials and 2 snack items is given by
f( x = 1, y = 2) =.05.
The probability of a customer purchasing 1 snack item only is given by f(x = 1, y = 0) = .40.
The probability f(x = 0, y = 0) = 0 because the point of sale terminal is only used when someone
makes a purchase.
b.
The marginal probability distribution of x along with the calculation of the expected value and
variance is shown below.
xf (x)
x - E(x)
(x - E(x))2
(x-E(x))2f (x)
x
f (x)
0
0.13
0
-1.14
1.2996
0.1689
1
0.60
0.6
-0.14
0.0196
0.0118
2
0.27
0.54
0.86
0.7396
0.1997
1.14
0.3804
E(x)
Var(x)
We see that E(x) = 1.14 snack items and Var(x) = .3804.
c.
The marginal probability distribution of y along with the calculation of the expected value and
variance is shown below.
y - E(y)
(y - E(y))2
0
-0.5
0.25
0.15
0.30
0.3
0.5
0.25
0.075
0.10
0.2
1.5
2.25
0.225
y
f (y)
0
0.60
1
2
yf (y)
0.5
(y-E(y))2f (y)
0.45
E(y)
Var(y)
We see that E(y) = .50 reading materials and Var(y) = .45.
5 - 30
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Discrete Probability Distributions
d.
The probability distribution of t = x + y is shown below along with the calculation of its expected
value and variance.
t
f (t)
tf (t)
t-E(t)
(t-E(t))2
(t-E(t))2f (t)
1
0.50
0.5
-0.64
0.4096
0.2048
2
0.38
0.76
0.36
0.1296
0.0492
3
0.10
0.3
1.36
1.8496
0.1850
4
0.02
0.08
2.36
5.5696
0.1114
1.64
0.5504
E(t)
Var(t)
We see that the expected number of items purchased is E(t) = 1.64 and the variance in the number of
purchases is Var(t) = .5504.
e.
From part (b), Var(x) = .3804. From part (c), Var(y) = .45. And from part (d), Var(x + y) = Var(t) =
.5504. Therefore,
 xy  [Var ( x  y )  Var ( x)  Var ( y )] / 2
 (.5504  .3804  .4500) / 2
 .14
To compute the correlation coefficient, we must first obtain the standard deviation of x and y.
 x  Var ( x)  .3804  .6168
 y  Var ( y )  .45  .6708
So the correlation coefficient is given by
 xy 
 xy
.14

 .3384
 x y (.6168)(.6708)
The relationship between the number of reading materials purchased and the number of snacks
purchased is negative. This means that the more reading materials purchased the fewer snack items
purchased and vice versa.
63. a.
b.
The All World stock fund would be considered the more risky because it has a larger standard
deviation. Indeed, the stock fund will experience a loss if the return is one standard deviation of
18.9% below the mean or 7.80%.
To answer this question we need to compute the expected value, variance, and standard deviation of
.75x + .25y.
E (.75x  .25 y)  .75E ( x)  .25E ( y)  .75(7.8)  .25(5.5)  7.225
To compute the variance and standard deviation of the portfolio, we need to first compute the
variance for x and y.
5 - 31
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Chapter 5
Var ( x)  18.92  357.21 and Var ( y)  4.62  21.16
Var (.75 x  .25 y )  .752Var ( x)  .252Var ( y )  2(.75)(.25) xy
 .752 (357.21)  .252 (21.16)  2(.75)(.25)(12.4)
 197.60
and  xy  197.6  14.06
Expected return ($10,000 investment) = 10,000(.07225) = $722.50
Standard deviation ($10,000 investment) = 10,000(.1406) = $1406
c.
To answer this question we need to compute the expected value, variance, and standard deviation of
.25x + .75y.
E (.25 x  .75 y )  .25 E ( x)  .75 E ( y )  .25(7.8)  .75(5.5)  6.075
Var (.25 x  .75 y )  .252Var ( x)  .752Var ( y )  2(.25)(.75)( 12.4)
 .252 (357.21)  .752 (21.16)  2(.25)(.75)(12.4)
 29.5781
 .25 x .75 y  29.5781  5.4386
Expected return ($10,000 investment) = 10,000(.06075) = $607.50
Standard deviation ($10,000 investment) = 10,000(.054386) = $543.86
d.
I would recommend the portfolio in part (b) for an aggressive investor because it has a larger return.
I would recommend the portfolio in part (c) for a conservative investor because it has a smaller
standard deviation and is, thus, less risky.
64. a.
n = 20
and
x = 3
 20 
f (3)    (.05)3 (.95)17  .0596
3
b.
n = 20
and
x = 0
 20 
f (0)    (.05)0 (.95) 20  .3585
0
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Discrete Probability Distributions
c.
E(x) = n p = 2000(.05) = 100
The expected number of employees is 100.
d.
 = np (1 - p) = 2000(.05)(.95) = 95
=
65. a.
95 = 9.75
We must have E(x) = np  20
For the 18-34 age group, p = .26.
n(.26)  20
n  76.92
Sample at least 77 people to have an expected number of home owners at least 20 for this age group.
b.
For the 35-44 age group, p = .50.
n(.50)  20
n  40
Sample at least 40 people to have an expected number of home owners at least 20 for this age group.
c. For the 55 and over age group, p = .88.
n(.88)  20
n  22.72
Sample at least 23 people to have an expected number of home owners at least 20 for this age group.
d.
  np(1  p)  77(.26)(.74)  3.85
e.
  np(1  p)  40(.50)(.50)  3.16
66.
Since the shipment is large we can assume that the probabilities do not change from trial to trial and
use the binomial probability distribution.
a.
n = 5
5
f (0)    (0.01)0 (0.99)5  .9510
0
b.
5
f (1)    (0.01)1 (0.99) 4  .0480
1
c.
1 - f (0) = 1 - .9510 = .0490
d.
No, the probability of finding one or more items in the sample defective when only 1% of the items
in the population are defective is small (only .0490). I would consider it likely that more than 1% of
the items are defective.
5 - 33
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Chapter 5
67. a.
b.
E(x) = np = 100(.041) = 4.1
Var(x) = np(1 - p) = 100(.041)(.959) = 3.93
  3.93  1.98
68. a.
E(x) = 800(.30) = 240
b.
  np(1  p)  800(.30)(.70)  12.96
c.
For this one p = .70 and (1-p) = .30, but the answer is the same as in part (b). For a binomial
probability distribution, the variance for the number of successes is the same as the variance for the
number of failures. Of course, this also holds true for the standard deviation.
 = 15
69.
P(20 or more arrivals) = f (20) + f (21) + · · ·
= .0418 + .0299 + .0204 + .0133 + .0083 + .0050 + .0029
+ .0016 + .0009 + .0004 + .0002 + .0001 + .0001 = .1249
 = 1.5
70.
P(3 or more breakdowns) = 1 - [ f (0) + f (1) + f (2) ].
1 - [ f (0) + f (1) + f (2) ]
= 1 - [ .2231 + .3347 + .2510]
= 1 - .8088 = .1912
 = 10 f (4) = .0189
71.
72. a.
b.
f (3) 
33 e3
 .2240
3!
f (3) + f (4) + · · · = 1 - [ f (0) + f (1) + f (2) ]
0
f (0) = 3 e
0!
-3
= e
-3
= .0498
Similarly, f (1) = .1494, f (2) = .2240
 1 - [ .0498 + .1494 + .2241 ] = .5767
73.
Hypergeometric N = 52, n = 5 and r = 4.
a.
4IF
48I
F
G
J
G
H2KH3 J
K 6(17296) .0399
52
F
IJ 2,598,960
G
H5 K
5 - 34
© 2013 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Discrete Probability Distributions
b.
c.
d.
74. a.
4IF
48I
F
G
J
G
H1KH4 J
K 4(194580) .2995
52I
2,598,960
F
G
J
H5 K
4IF
48I
F
G
J
G
H0KH5 J
K 1,712,304 .6588
52
F
IJ 2,598,960
G
H5 K
1 - f (0) = 1 - .6588 = .3412
Hypergeometric distribution with N = 10, n =2, and r = 7.
 r  N  r 
 

x nx 
f (1)   

N
 
n
 7  10  7   7  3 
 
   
 1  2  1    1  1   (7)(3)  .4667
45
10 
 10 
 
 
2
2
b.
 7 3
  
2 0
(21)(1)
f (2)      
 .4667
45
 10 
 
2
c.
 7  3 
  
0 2
(1)(3)
f (0)     
 .0667
45
10 
 
2
5 - 35
© 2013 Cengage Learning. All Rights Reserved.
May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.