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Chapter 5:
Continuous Random Variables and Probability
Distributions
5.1 P(1.4 < X < 1.8) = F(1.8) – F(1.4) = (.5)(1.8) – (.5)(1.4) = 0.20
5.2 P(1.0 < X < 1.9) = F(1.9) – F(1.0) = (.5)(1.9) – (.5)(1.0) = 0.45
5.3 P(X < 1.4) = F(1.4) = (.5)(1.4) = 0.7
5.4 P(X > 1.3) = F(1.3) = (.5)(2.0) – (.5)(1.3) = 0.35
5.5 a.
Probability Density Function: f(x)
1.5
f(x)
1.0
0.5
0.0
0
X
1
Chapter 5: Continuous Random Variables and Probability Distributions
b.
Cumulative distribution function: F(x)
F(x)
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
C17
c. P(X < .25) = .25
d. P(X >.75) = 1-P(X < .75) = 1-.75 = .25
e. P(.2 < X < .8) = P(X <.8) – P(X <.2) = .8 - .2 = .6
5.6
a.
Probability density function: f(x)
0.75
f(x)
0.50
0.25
0.00
0
1
2
X
3
4
123
124
Statistics for Business & Economics, 7th edition
b.
Cumulative density function: F(x)
1.0
F(x)
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
X
c. P(x < 1) = .25
d. P(X < .5) + P(X > 3.5)=P(X < .5) + 1 – P(X < 3.5) = .25
5.7
a. P(60,000 < X< 72,000) = P(X < 72,000) – P(X < 60,000) = .6 - .5 = .1
b. P(X < 60,000) < P(X < 65,000) < P(X < 72,000); .5 < P(X < 65,000) < .6
5.8
a. P(380 < X < 460) = P(X < 460) – P(X < 380) = .6 - .4 = .2
b. P(X < 380) < (PX< 400) < P(X < 460); .4 < P(X < 400) < .6
5.9
W = a + bX. If TC = 1000 + 2X where X = number of units produced,
find the mean and variance of the total cost if the mean and variance for
the number of units produced are 500 and 900 respectively. W  a  b x
= 1000 + 2(500) = 2000.  2W  b 2 2 X = (2)2(900) = 3600.
5.10
W = a + bX. If Available Funds = 1000 - 2X where X = number of units
produced, find the mean and variance of the profit if the mean and
variance for the number of units produced are 50 and 90 respectively.
W  a  b x = 1000 - 2(50) = 900.  2W  b 2 2 X = (-2)2(90) = 360.
5.11
W = a + bX. If Available Funds = 2000 - 2X where X = number of units
produced, find the mean and variance of the profit if the mean and
variance for the number of units produced are 500 and 900 respectively.
W  a  b x = 2000 - 2(500) = 1000.  2W  b 2 2 X = (-2)2(900) = 3600.
5.12
W = a + bX. If Available Funds = 6000 - 3X where X = number of units
produced, find the mean and variance of the profit if the mean and
variance for the number of units produced are 1000 and 900 respectively.
W  a  b x = 6000 - 2(1000) = 4000.  2W  b 2 2 X = (-3)2(900) = 8100
Chapter 5: Continuous Random Variables and Probability Distributions
5.13
Y = 10,000 + 1.5 X = 10,000 + 1.5 (30,000) = $55,000
Y = |1.5| X = 1.5 (8,000) = $12,000
5.14
Y = 20 + X = 20 + 4 = $24 million
Bid = 1.1 Y =1.1(24) = $26.4 million,  = $1 million
5.15
Y = 60 + .2 X = 60 + 140 = $200
Y = |.2| X = .2 (130) = $26
5.16
Y = 6,000 + .08 X = 6,000 + 48,000 = $54,000
Y = |.08| X = .08(180,000) = $14,400
5.17
a.
b.
c.
d.
e.
f.
g.
P(Z < 1.20) = .8849
P(Z > 1.33) = 1 – Fz(1.33) = 1 - .9082 = .0918
P(Z < -1.70) = 1 – Fz(1.70) = 1 - .9554 = .0446
P(Z > -1.00) = Fz(1) = .8413
P(1.20 < Z < 1.33) = Fz(1.33) – Fz(1.20) = .9082 - .8849 = .0233
P(-1.70 < Z < 1.20) = Fz(1.20) – [1 - Fz(1.70)] = .8849 – .0446 = .8403
P(-1.70 < Z < -1.00) = Fz(1.70) – Fz(1.00) = .9554 - .8413 = .1141
5.18
a.
b.
c.
d.
Find Z0 such that P(Z < Z0) = .7, closest value of Z0 = .52
Find Z0 such that P(Z < Z0) = .25, closest value of Z0 = -.67
Find Z0 such that P(Z > Z0) = .2, closest value of Z0 = .84
Find Z0 such that P(Z > Z0) = .6, closest value of Z0 = -.25
5.19
X follows a normal distribution with µ = 50 and 2 = 64
60  50
a. Find P(X > 60). P(Z >
) = P(Z > 1.25)
8
= .5 - .3944 = .1056
35  50
62  50
b. Find P(35 < X < 62). P(
<Z<
) = P(-1.88 < Z < 1.5)
8
8
= .4699 + .4332 = .9031
55  50
c. Find P(X < 55). P(Z <
) = P(Z < .62)
8
= .5 + .2324 = .7324
d. Probability is .2 that X is greater than what number? Z = .84.
X  50
.84 
X = 56.72
8
e. Probability is .05 that X is in the symmetric interval about the mean
X  50
between? Z = +/- .06. .06 
. X = 49.52 and 50.48.
8
125
126
Statistics for Business & Economics, 7th edition
5.20
X follows a normal distribution with µ = 80 and 2 = 100
60  80
a. Find P(X > 60). P(Z >
) = P(Z > -2.00) = .5 + .4772 = .9772
10
72  80
82  80
b. Find P(72 < X < 82). P(
<Z<
) = P(-.80 < Z < .20) =
10
10
.2881 + .0793 = .3674
55  80
c. Find P(X < 55). P(Z <
) = P(Z < -2.50) = .5 - .4938 = .0062
10
d. Probability is .1 that X is greater than what number? Z = 1.28.
X  80
1.28 
X = 92.8
10
e. Probability is .08 that X is in the symmetric interval about the mean
X  80
between? Z = +/- .10. .10 
. X = 79 and 81.
10
5.21
X follows a normal distribution with µ = .2 and 2 = .0025
.4  .2
a. Find P(X > .4). P(Z >
) = P(Z > 4.00) = .5 - .5 = .0000
.05
.15  .2
.28  .2
b. Find P(.15 < X < .28). P(
<Z<
) = P(-1.00 < Z < 1.60)
.05
.05
= .3431 + .4452 = .7883
.10  .20
c. Find P(X < .10). P(Z <
) = P(Z < -2.00) = .5 - .4772 = .0228
.05
d. Probability is .2 that X is greater than what number? Z = .84.
X  .20
.84 
X = .242
.05
e. Probability is .05 that X is in the symmetric interval about the mean
X  .2
between? Z = +/- .06. .06 
. X = .197 and .203.
.05
5.22
a. P(Z <
b.
c.
d.
e.
400  380
) = P(Z < .4) = .6554
50
360  380
P(Z >
) = P(Z > -.4) = FZ(.4) = .6554
50
The graph should show the property of symmetry – the area in the tails
equidistant from the mean will be equal.
300  380
400  380
P(
<Z<
) = P(-1.6 < Z < .4) = FZ(.4) – [150
50
FZ(1.6)] = .6554 - .0548 = .6006
The area under the normal curve is equal to .8 for an infinite number of
ranges – merely start at a point that is marginally higher. The shortest
range will be the one that is centered on the z of zero. The z that
Chapter 5: Continuous Random Variables and Probability Distributions
corresponds to an area of .8 centered on the mean is a Z of ±1.28. This
yields an interval of the mean plus and minus $64: [$316, $444]
1, 000  1, 200
) = P(Z > -2) =FZ(2) = .9772
100
1,100 1,200
1,300  1, 200
b. P(
<Z<
) = P(-1 < Z < 1) = 2FZ(1) –1 = .6826
100
100
c. P(Z > 1.28) = .1, plug into the z-formula all of the known information
Xi  1, 200
and solve for the unknown: 1.28 =
. Solve algebraically
100
for Xi = 1,328
5.23
a. P(Z >
5.24
a. P(Z >
5.25
a. P(Z >
5.26
a. P(Z <
38  35
) = P(Z > .75) = 1 - FZ(.75) = .2266
4
32  35
b. P(Z <
) = P(Z < -.75) = 1 - FZ(.75) = .2266
4
32  35
38  35
c. P(
<Z<
) = P(-.75 < Z < .75) = 2FZ(.75) – 1 =
4
4
2(.7734) – 1 = .5468
d. (i) The graph should show the property of symmetry – the area in the
tails equidistant from the mean will be equal.
(ii) The answers to a, b, c sum to one because the events cover the
entire area under the normal curve which by definition, must sum to 1.
20  12.2
) = P(Z > 1.08) = 1 – Fz (1.08) = .1401
7.2
0  12.2
b. P(Z <
) = P(Z < -1.69) = 1 – Fz (1.69) = .0455
7.2
5  12.2
15  12.2
c. P(
<Z<
) = P(-1 < Z < .39) = Fz (.39) – [1- Fz (1)] =
7.2
7.2
.6517 - .1587 = .4930
10  12.2
) = P(Z < - .79) = 1 – Fz (.79) = .2148
2.8
15  12.2
b. P(Z >
) = P(Z > 1) = 1 – Fz (1) = .1587
2.8
12  12.2
15  12.2
c. P(
<Z<
) = P(-.07 < Z < 1) = Fz (1) – [1- Fz (.07)]
2.8
2.8
= .8413 - .4721 = .3692
d. The answer to a. will be larger because 10 grams is closer to the mean
than is 15 grams. Thus, there would be a greater area remaining less
than 10 grams than will be the area above 15 grams.
127
128
Statistics for Business & Economics, 7th edition
460  500
540  500
<Z<
) = P(-.8 < Z < .8) = 2 Fz (.8) – 1 = .5762
50
50
b. If P(Z < -.84) = .2, then plug into the z formula and solve for the Xi:
Xi  500
the value of the cost of the contract. -.84 =
. Xi = $458
50
(thousand dollars)
c. The shortest 95% range will be the interval centered on the mean.
Xi  500
Since the P(Z > 1.96) = .025, 1.96 =
. Xi = 598. The
50
Xi  500
lower value of the interval will be –1.96 =
which is Xi =
50
$402 (thousand dollars). Therefore, the shortest range will be 598 –
402 = $196 (thousand dollars).
5.27
a. P(
5.28
P(Z > 1.5) = 1 - Fz(1.5) = .0668
5.29
P(Z < -1.28) = .1, –1.28 =
5.30
P(Z > .67) = .25, .67 = 17.8 - 
P(Z > 1.03) = .15, 1.04 = 19.2 - 
Solving for , :  = 15.265, 2 = (3.7838)2 = 14.317
5.31
a. P(Z >
5.32
For Investment A, the probability of a return higher than 10%:
10  10.4
P(Z >
) = P(Z > -.33) = FZ(.33) = .6293
1.2
For Investment B, the probability of a return higher than 10%
10  11.0
P(Z >
) = P(Z > -.25) = FZ(.25) = .5987
4
Therefore, Investment A is a better choice
Xi  18.2
Xi = 16.152
1.6
820  700
) = P(Z> 1) = 1 – Fz (1) = .1587
120
730  700
820  700
b. P(
<Z<
) = P(.25 < Z < 1) = .8413 - .5987 =
120
120
.2426
Number of students = .2426(100) = 24.26 or 24 students
Xi  700
c. P(Z < -1.645) = .05, –1.645 =
, Xi = 502.6
120
Chapter 5: Continuous Random Variables and Probability Distributions
5  4.4
) = P(Z < 1.5) = .9332
.4
5  4.2
For Supplier B: P(Z <
) = P(Z < 1.33) = .9082
.6
Therefore, Supplier A has a greater probability of achieving less than 5%
impurity and is hence the better choice
5.33
For Supplier A: P(Z <
5.34
a. P(Z > -1.28) = .9, -1.28 =
5.35
a. P(Z <
5.36
a. P(
5.37
a. P(
Xi  150
, Xi = 98.8
40
Xi  150
b. P(Z < .84) = .8, .84 =
, Xi = 183.6
40
120  150 2
c. P(X  1) = 1 – P(X = 0) = 1-[P(Z<
)] = 1 – [P(Z < -.75)]2 =
40
1 – (.2266)2 = .9487
60  75
) = P(Z < -.75) = .2266
20
90  75
b. P(Z >
) = P(Z >.75) = .2266
20
c. The graph should show that 60 minutes and 90 minutes are equidistant
from the mean of 75 minutes. Therefore, the areas above 90 minutes
and below 60 minutes by the property of symmetry must be equal.
Xi  75
d. P(Z > 1.28) = .1, 1.28 =
, Xi = 100.6
20
400  420
480  420
<Z<
) = P(-.25 < Z < .75) = Fz (.75) – [1 – FZ
80
80
(.25)] = .7734 - .4013 =.3721
Xi  420
b. P(Z > 1.28) = .1, 1.28 =
, Xi = 522.4
80
c. 400 – 439
d. 520 – 559
500  420 2
e. P(X 1) = 1 –P(X = 0 ) = 1 – [P(Z<
)] = 1 – (.8413)2 = .2922
80
180  200
< Z < 0) = .5 – [1- Fz (1)] = .5 -.1587 = .3413
20
245  200
b. P(Z >
) = 1 – FZ(2.25) = .0122
20
c. Smaller
Xi  200
d. P(Z < -1.28) = .1, -1.28 =
, Xi = 174.4
20
129
130
5.38
Statistics for Business & Economics, 7th edition
P(Z < 1.5) = .9332, 1.5 =
85  70

,  = 10
80  70
) = P(Z > 1) = .1587
10
P(X  1) = 1 – P(X=0) = 1 – [FZ(1)]4 = 1 – (.8413)4 = .4990
P(Z >
5.39
n = 900 from a binomial probability distribution with P = .50
a. Find P(X > 500). E[X] =  = 900(.5) = 450,  = (900)(.5)(.5) = 15
500  450
P(Z >
) = P(Z > 3.33) = 1 – FZ(3.33) = .0004
15
430  450
b. Find P(X < 430). P(Z <
) = P(Z < -1.33) = 1 - FZ(1.33) = .0918
15
440  450
480  450
c. P(
<Z<
) = P(-.67 < Z < 2.00) = fz (-.67) +
15
15
fZ(2.00) = .2486 + .4772 = .7258
d. Probability is .1 that the number of successes is less than how many?
X  450
Z=
-1.28. 1.28 
X = 430.8
15
e. Probability is .08 the number of successes is greater than? Z = 1.41.
X  450
1.41 
. X = 471.15.
15
5.40
n = 1600 from a binomial probability distribution with P = .40
a. Find P(X > 1650). E[X] =  = 1600(.4) =
640,  = (1600)(.4)(.6) = 19.5959 P(Z >
1650  1600
) = P(Z > 2.55) = 1 – FZ(2.55) =
19.5959
.0054
1530  1600
b. Find P(X < 1530). P(Z <
) = P(Z
19.5959
< -3.57) = 1 - FZ(3.57) = .0002
1550  1600
1650  1600
c. P(
<Z<
) = P(-2.55
19.5959
19.5959
< Z < 2.55) = (2)Fz (2.55) = (2).4946 =
.9892
d. Probability is .09 that the number of successes is less
X  1600
than how many? Z =
-1.34. 1.34 
X
19.5959
= 1573.741 1,574 successes
Chapter 5: Continuous Random Variables and Probability Distributions
e.
Probability is .20 the number of successes is greater
X  1600
than? Z = .84. .84 
. X = 1616.46
19.5959
1,616 successes
5.41
n = 900 from a binomial probability distribution with P = .10
a. Find P(X > 110). E[X] =  = 900(.1) = 90, 
110  90
= (900)(.1)(.9) = 9
P(Z >
)
9
= P(Z > 2.22) = 1 – FZ(2.22) = .0132
53  90
b. Find P(X < 53). P(Z <
) = P(Z < 9
4.11) = 1 - FZ(4.11) = .0000
55  90
120  90
c. P(
<Z<
) = P(-3.89 < Z <
9
9
3.33) = 1.0000
d. Probability is .10 that the number of successes
is less than how many? Z =
-1.28.
X  90
1.28 
X = 78.48
9
e. Probability is .08 the number of successes is
X  90
greater than? Z = 1.41. 1.41 
. X=
9
102.69
5.42
n = 1600 from a binomial probability distribution with P = .40
a. Find P(P > .45). E[P] =  = P = .40,  =
P(1  P)
.4(1  .4)
= .01225 P(Z >

n
1600
.45  .40
) = P(Z > 4.082) = 1 – FZ(4.082) =
.01225
.0000
.36  .40
b. Find P(P < .36). P(Z <
) = P(Z < .01225
3.27) = 1 - FZ(3.27) = .0005
.44  .40
.37  .40
c. P(
<Z<
) = P(3.27 < Z < .01225
.01225
2.45) = 1 – [(2)[1-Fz (3.27)]] = 1 - (2)[1.9995] = .9995 - .0071 = .9924
d. Probability is .20 that the percentage of
successes is less than what percent? Z = -.84.
X  .40
.84 
P = 38.971%
.01225
131
132
Statistics for Business & Economics, 7th edition
e. Probability is .09 the percentage of successes
X  .40
is greater than? Z = 1.34. 1.34 
. P=
.01225
41.642%
Chapter 5: Continuous Random Variables and Probability Distributions
5.43
5.44
5.45
5.46
n = 400 from a binomial probability distribution with P = .20
a. Find P(P > .25). E[P] =  = P = .20,  =
P(1  P)
.2(1  .8)
= .02
P(Z >

n
400
.25  .20
) = P(Z > 2.50) = 1 – FZ(2.50) = 1 .02
.4938 = .0062
.16  .20
b. Find P(P < .16). P(Z <
) = P(Z < .02
2.00) = 1 - FZ(2.00) = .0228
.17  .20
.24  .20
c. P(
<Z<
) = P(-1.50 < Z <
.02
.02
2.00) = [Fz (1.50) - .5] + [Fz (2.00) - .5] =
.4332 + .4772 = .9104
d. Probability is .15 that the percentage of
successes is less than what percent? Z = -1.04.
X  .20
1.04 
P = 17.92%
.02
e. Probability is .11 the percentage of successes
X  .20
is greater than? Z = 1.23. 1.23 
. P=
.02
22.46%
a. E[X] =  = 900(.2) = 180,  = (900)(.2)(.8) = 12
200  180
P(Z >
) = P(Z > 1.67) = 1 – FZ(1.67) = .0475
12
175  180
b. P(Z <
) = P(Z < -.42) = 1 - FZ(.42) = .3372
12
a. E[X] =  = 400(.1) = 40,  = (400)(.1)(.9) = 6
35  40
P(Z >
) = P(Z > -.83) = FZ(.83) = .7967
6
40  40
50  40
b. P(
<Z<
) = P(0 < Z < 1.67) = Fz (1.67) – FZ(0) = 9525
6
6
- .5 = .4525
34  40
48  40
c. P(
<Z<
) = P(-1 < Z < 1.33) = Fz (1.33) – [1 – FZ(1)]
6
6
= .9082 - .1587 = .7495
d. 39 - 41
E[X] = (100)(.6) = 60,  =
(100)(.6)(.4) = 4.899
133
134
Statistics for Business & Economics, 7th edition
P(Z <
50  60
) = P(Z < -2.04) = 1 – FZ(2.04) = 1- .9793 = .0207
4.899
Chapter 5: Continuous Random Variables and Probability Distributions
5.47
5.48
5.49
5.50
a. E[X] = (450)(.25) = 112.5,  = (450)(.25)(.75) = 9.1856
100  112.5
P(Z <
) = P(Z < -1.36) = 1 - FZ(1.36) = 1 - .9131 = .0869
9.1856
120  112.5
150  112.5
b. P(
<Z<
) = P(.82 < Z < 4.08) = Fz(4.08) 9.1856
9.1856
Fz(.82) = 1.000 - .7939 = .2061
38  35
) = P(Z > .75) = 1 - FZ(.75) = 1 - .7734 = .2266
4
E[X] = 100(.2266) = 22.66,  = (100)(.2266)(.7734) = 4.1863
25  22.66
P(Z >
) = P(Z > .56) = 1 - FZ(.56) = 1 - .7123 = .2877
4.1863
P(Z >
10  12.2
) = P(Z < -.79) = 1 - FZ(.79) = 1 - .7852 = .2148
2.8
E[X] = 400(.2148) = 85.92,  = (400)(.2148)(.7852) = 8.2137
100  85.92
P(Z >
) = P(Z > 1.71) = 1 - FZ(1.71) = 1 - .9564 = .0436
8.2137
P(Z ≤
 = 1.0, what is the probability that an arrival occurs in the first t=2 time
units?
Cumulative Distribution Function
Exponential with mean = 1
x P( X <= x )
0
0.000000
1
0.632121
2
0.864665
3
0.950213
4
0.981684
5
0.993262
P(T < 2) = .864665
5.51  = 8.0, what is the probability that an arrival occurs in the first t=7 time
units?
Cumulative Distribution Function
Exponential with mean = 8
x P( X <= x )
0
0.000000
1
0.117503
2
0.221199
3
0.312711
4
0.393469
5
0.464739
6
0.527633
7
0.583138
8
0.632121
P(T < 7) = .583138
135
136
5.52
Statistics for Business & Economics, 7th edition
 = 5.0, what is the probability that an arrival occurs after t=7 time units?
Cumulative Distribution Function
Exponential with mean = 5
x P( X <= x )
0
0.000000
1
0.181269
2
0.329680
3
0.451188
4
0.550671
5
0.632121
6
0.698806
7
0.753403
8
0.798103
P(T>7) = 1-[P(T ≤ 8)] = 1 - .7981 = .2019
5.53
 = 6.0, what is the probability that an arrival occurs after t=5 time units?
Cumulative Distribution Function
Exponential with mean = 6
x P( X <= x )
0
0.000000
1
0.153518
2
0.283469
3
0.393469
4
0.486583
5
0.565402
6
0.632121
P(T>5) = 1-[P(T≤6)] = 1 - .6321 = .3679
5.54
 = 3.0, what is the probability that an arrival occurs after t=2 time units?
Cumulative Distribution Function
Exponential with mean = 3
x P( X <= x )
0
0.000000
1
0.283469
2
0.486583
3
0.632121
P(T<2) = .4866
5.55 a. P(X < 20) = 1 - e  (20 /10) = .8647
b. P(X > 5) = 1 – [1 - e  (5 /10) ] = e  (5 /10) = .6065
c. P(10 < X < 15) = (1- e  (15 /10) - (1 - e  (10 /10) ) = e 1 - e 1.5 = .1447
5.56 P(X > 18) = e  (18 /15) = .3012
5.57 P(X > 2) = e  (2)(.8) = .2019
5.58 a. P(X > 3) = 1 – [1 - e (3/  ) ] = e 3 since  = 1 / 
b. P(X > 6) = 1 – [1 - e (6 /  ) ] = e (6 /  ) = e 6 
c. P(X>6|X>3) = P(X > 6)/P(X > 3) = e 6  / e 3 ] = e 3
The probability of an occurrence within a specified time in the future is
not related to how much time has passed since the most recent
occurrence.
Chapter 5: Continuous Random Variables and Probability Distributions
5.59
Find P (t  3) . Note that   40 calls / 60 minutes  2 calls / 3 minutes.
P(t  3)  1  P(t  3)  1  [1  e ( 2 / 3)(3) ]  e2  0.1353
5.60
Let   20 trucks / 60 minutes  1 truck / 3 minutes.
a. P(t  5)  1  P(t  5)  1  [1  e (1/ 3)(5) ]  0.1889
b. P(t  1)  1  e(1/ 3)( 2)  0.4866
c. P(4  t  10)  [1  e (1/ 3)(10) ]  [1  e (1/ 3)( 4) ]  0.2279
5.61
Find the mean and variance of the random variable: W = 5X + 4Y with
correlation = .5
W  a  x  b y = 5(100) + 4(200) = 1300
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 52(100) + 42(400) + 2(5)(4)(.5)(10)(20) = 12,900
5.62
Find the mean and variance of the random variable: W = 5X + 4Y with
correlation = -.5
W  a  x  b y = 5(100) + 4(200) = 1300
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 52(100) + 42(400) + 2(5)(4)(-.5)(10)(20) = 4,900
5.63 Find the mean and variance of the random variable: W = 5X – 4Y with
correlation = .5.
W  a  x  b y = 5(100) – 4(200) = -300
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 52(100) + 42(400) – 2(5)(4)(.5)(10)(20) = 4900
5.64 Find the mean and variance of the random variable: W = 5X – 4Y with
correlation = .5.
W  a  x  b y = 5(500) – 4(200) = 1700
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 52(100) + 42(400) – 2(5)(4)(.5)(10)(20) = 4900
5.65 Find the mean and variance of the random variable: W = 5X – 4Y with
correlation of -.5.
W  a  x  b y = 5(100) – 4(200) = -300
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 52(500) + 42(400) – 2(5)(4)(-.5)(22.3607)(20) = 27,844.28
137
138
Statistics for Business & Economics, 7th edition
5.66
 Z = 100,000(.1) + 100,000(.18).  x = 10,000 + 18,000 = 28,000
Z = 0. Note that the first investment yields a certain profit of 10%
which is a zero standard deviation. x = 100,000(.06) = 6,000
5.67
Assume that costs are independent across years
 Z = 5  x = 5(200) = 1,000
Z =
5.68
5 x =
2
5(3,600) = 134.16
 Z = 1  2  3 = 50,000 + 72,000 + 40,000 = 162,000
Z =  1   2   3 =
2
5.69
2
2
 Z = 1  2  3 = 20,000 + 25,000 + 15,000 = 60,000
Z =  1   2   3 =
2
5.70
(10, 000) 2  (12, 000) 2  (9, 000) 2 = 18,027.76
2
2
(2, 000) 2  (5, 000) 2  (4, 000) 2 = 6,708.2
The calculation of the mean is correct, but the standard deviations of two
random variables cannot be summed. To get the correct standard
deviation, add the variances together and then take the square root. The
standard deviation:   5(16) 2 = 35.7771
5.71
 Z = (16  x ) / 16 =  x = 28
Z = 16 x / 16 =
2
5.72
(2.4) 2 / 16 = 2.4 / 4 = .6
a. Compute the mean and variance of the portfolio with correlation of +.5
W  a  x  b y = 50(25) + 40(40) = 2850
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 502(121) + 402(225) + 2(50)(40)(.5)(11)(15) = 992,500
b. Recompute with correlation of -.5
W  a  x  b y = 50(25) + 40(40) = 2850
5.73
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 502(121) + 402(225) + 2(50)(40)(-.5)(11)(15) = 332,500
a. Find the probability that total revenue is greater than total cost
W = aX – bY = 10X –[7Y+25)]
W  a  x  b y = 10(100) – [7(100) + 250] = 50
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 102(64) + 72(625) – 2(10)(7)(.6)(8)(25) = 20,225 W  20, 225
= 142.2146
Chapter 5: Continuous Random Variables and Probability Distributions
P(Z >
0  50
) = P(Z > -.35) = FZ(.35) = .6368
142.2146
b. 95% acceptance interval = 50 ± 1.96 (142.2146) = 50 ± 278.7406 = 228.7406 to 328.7406
5.74
a. W = aX – bY = 10X – 10Y
W  a  x  b y = 10(100) – 10(90) = 100
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
=102(100) + 102(400) – 2(10)(10)(-.4)(10)(20) =66,000 W  66,000
=256.90465
b. P(Z <
0  100
) = P(Z < -.39) = 1 – FZ(.39) = 1 – .6517 = .3483
256.90465
5.75
W = aX – bY = 10X – 4Y
W  a  x  b y = 10(400) – 4(400) = 2400
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
=102(900) + 42(1600) – 2(10)(4)(.5)(30)(40) = 67,600 W  67,600 =260
P(Z >
5.76
2000  2400
) = P(Z > -1.54) = FZ(1.54) = .9382
260
a. W = aX – bY = 1X – 1Y
W  a  x  b y = 1(100) – 1(105) = -5
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
=12(900) + 12(625) – 2(1)(1)(.7)(30)(25) = 475 W  475 =21.79449
b. P(Z >
5.77
a.
b.
c.
d.
0  ( 5)
) = P(Z > .23) = 1 – FZ(.23) = 1 – .5910 = .4090
21.79449
P(X < 10) = (10/12) – (8/12) = 1/6
P(X > 12) = (20/12) – (12/12) = 8/12 = 2/3
E[] = (20/12 – 12/12) = 2(2/3) = 1.333
To jointly maximize the probability of getting the contract and the
profit from that contract, maximize the following function: max E[]
= (B – 10)(20/12 – B/12). Where B is the value of the bid. To
determine the value for B that maximizes the expected profit, an
iterative approach can be used. The value of B is 15.
139
140
Statistics for Business & Economics, 7th edition
5.78 a.
Probability Density Function: f(x)
f(x)
0.033333
0.000000
35
30
40
50
45
55
60
70
65
X
b. Cumulative density function
Cumulative density function: F(x)
1.0
F(x)
0.8
0.6
0.4
0.2
0.0
35
40
45
50
55
X
c. P(40 < X < 50) = (50/30) – (40/30) = 10/30
65  35
d. E[X] =
= 50
2
60
65
Chapter 5: Continuous Random Variables and Probability Distributions
5.79
a. The probability density function f(x):
Probability density function: f(x)
1.50
f(x)
1.00
0.5
0.00
0
.5
1
1.5
3
X
b. Fx(x)  0 for all x. The area under fx(x) = 2[½(base x height)] = 1
.52
.52
c. P(.5 < X < 1.5 ) = (.5 ) + (.5 ) = .375 + .375 = .75
2
2
5.80
5.81
a.  Y = 2000(1.1) + 1000(1+  x ) = 2,200 + 1,160 = 3,360
b. Y = |1000| x = 1000(.08) = 80
a.  R = 1.45  x = 1.45(530) = 768.5
b. R = |1.45| x = 1.45(69) = 100.05
c.  = R – C = .5X – 100, E[] = .5  x -100 = 165,  = |.5| x = .5(69) = 34.5
5.82
Given that the variance of both predicted earnings and forecast error are
both positive and given that the variance of actual earnings is equal to the
sum of the variances of predicted earnings and forecast error, then the
Variance of predicted earnings must be less than the variance of actual
earnings
5.83
Cov[(X1 + X2), (X1 – X2)] = E[(X1 + X2)(X1 – X2)] – E[X1 + X2] E[X1 –
X2] = E[X12 - X22]– E[(X1) + E(X2)][E(X1) – E(X2)] =
E(X12) – E(X22) - [(E(X1))2 – (E(X2)2] = Var (X1) – Var (X2)
Which is 0 if and only if Var (X1) = Var (X2)
141
142
5.84
Statistics for Business & Economics, 7th edition
3  2.6
) = P(Z > .8) = 1 – FZ(.8) = .2119
.5
2.25  2.6
2.75  2.6
P(
<Z<
) = P(-.7 < Z < .3) = Fz (.3) – [1-FZ(.3)] =
.5
.5
.3759
Xi  2.6
P(Z > 1.28) = .1, 1.28 =
, Xi = 3.24
.5
P(Xi > 3) = .2119 (from part a)
E[X] = 400(.2119) = 84.76, x = (400)(.2119)(.7881) = 8.173
80  84.76
P(Z >
) = P(Z > -.58) = FZ(.58) = .7190
8.173
P(X  1) = 1 – P(X = 0) = 1 – (.7881)2 = .3789
a. P(Z >
b.
c.
d.
e.
65  60
) = P(Z > .5) = 1 – FZ(.5) = .3085
10
50  60
70  60
b. P(
<Z<
) = P(-1 < Z < 1) = 2 Fz (1) – 1 = .6826
10
10
Xi  60
c. P(Z > 1.96) = .025, 1.96 =
, Xi = 79.6
10
d. P(Z > .675) = .025, .675 = The shortest range will be the interval
Xi  60
centered on the mean. Since the P(Z > .675) = .025, .675 =
.
10
Xi  60
Xi = 66.75. The lower value of the interval will be –.675 =
10
which is Xi = 53.25. Therefore, the shortest range will be 66.75 –
53.25 = 13.5. This is by definition the InterQuartile Range (IQR).
P(X > 65) = .3085 (from part a)
Use the binomial formula: P(X = 2) = C24 (.3085)2 (.6915)2 = 0.2731
5.85
a. P(Z >
5.86
a. P(Z <
b.
c.
d.
e.
f.
g.
85  100
) = P(Z < -.5) = .3085
30
70  100
130  100
P(
<Z<
) = P(-1 < Z < 1) = 2 Fz (1) – 1 = .6826
30
30
Xi  100
P(Z > 1.645) = .05, 1.645 =
, Xi = 149.35
30
60  100
P(Z >
) = P(Z > -1.33) = FZ(1.33) = .9032
30
P(X  1) = 1 – P(X = 0) = 1 – (.0918)2 = .9916
Use the binomial formula: P(X = 2) = C24 (.9082)2 (.0918)2 = 0.0417
90 – 109
130 - 149
Chapter 5: Continuous Random Variables and Probability Distributions
5.87
b.
c.
d.
e.
f.
5.88
15  20
25  20
<Z<
) = P(-1.25 < Z < 1.25) = 2 FZ(1.25) – 1 =
4
4
.7888
30  20
P(Z >
) = P(Z > 2.5) =1 - Fz (2.5) = .0062
4
P(X  1) = 1 – P(X = 0) = 1 – [FZ(2.5)]5 = .0306
Xi  20
P(Z > .525) = .3, .525 =
, Xi = 22.1 The shortest range will be
4
the interval centered on the mean. The lower value of the interval will
Xi  20
be –.525 =
which is Xi = 17.9. Therefore, the shortest range
4
is 22.1 – 17.9 = 4.2.
19 – 21
21 – 23
a. P(
P(Z > 1.28) = .1, 1.28 =
P(Z >
5.89

,  = 23.4375
140  100
) = P(Z > 1.71) = 1 – FZ(1.71) = .0436
23.4375
P(Z > 1.28) = .1, 1.28 =
P(Z <
130  100
25  
,  = 21.8
2.5
20  21.8
) = P(Z < -.72) = 1 – FZ(.72) = .2358
2.5
5.90
E[X] = 1000(.4) = 400, x = (1000)(.4)(.6) = 15.4919
500  400
P(Z <
) = P(Z < 6.45)  1.0000
15.4919
5.91
E[X] = 400(.6) = 240, x = (400)(.6)(.4) = 9.798
200  240
P(Z >
) = P(Z > -4.08)  1.0000
9.798
5.92 The number of calls per 12-hour time period follows a Poisson distribution
with   15 calls / 12 - hour time period.
Cumulative Distribution Function
Poisson with mu = 15.0000
x
0.00
1.00
P( X <= x)
0.0000
0.0000
143
144
Statistics for Business & Economics, 7th edition
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
15.00
16.00
17.00
0.0000
0.0002
0.0009
0.0028
0.0076
0.0180
0.0374
0.0699
0.1185
0.1848
0.2676
0.3632
0.4657
0.5681
0.6641
0.7489
P( x  10)  P( x  9)  0.0699
P( x  17)  1  P( x  17)  1  0.7489  0.2511
5.93
e6 66
a. P(X = 6) =
= .1606
6!
b. 20 minutes = 1/3 hours, P(X > 1/3) = e

6
3
= .1353

6
c. 5 minutes = 1/12 hour, P(X < 1/12) = 1 - e 12 = .3935
d. 30 minutes = .5 hour, P(X > .5) = e  (.5)(6) = .0498
5.94
a. E[X] = 600(.4) = 240, x = (600)(.4)(.6) = 12
260  240
P(Z >
) = P(Z > 1.67) = 1 – FZ(1.67) = .0475
12
Xi  240
b. P(Z > -.254) = .6, -.254 =
, Xi = 236.95 (237 listeners)
12
5.95
a. P(
120  132
150  132
<Z<
) = P(- 1 < Z < 1.5) = FZ (1.5) – [1 – FZ(1)]
12
12
= .7745
Xi  132
b. P(Z > .44) = .33, .44 =
, Xi = 137.28
12
120  132
c. P(Z <
) = P(Z < -1) = 1 – FZ(1) = .1587
12
d. E[X] = 100(.1587) = 15.87, x = (100)(.1587)(.8413) = 3.654
25  15.87
P(Z >
) = P(Z > 2.5) = 1 – FZ(2.5) = .0062
3.654
Chapter 5: Continuous Random Variables and Probability Distributions
5.96
P(Z>1.28)=.1, 1.28=
3.5  2.4
3+ hours on task: P(Z >
400(.242) = 96.8, x =

, =.8594. Probability that 1 exec spends
3  2.4
) = P(Z > .7) = 1 – FZ(.7) = .242. E[X] =
.8594
80  96.8
)=P(Z>(400)(.242)(.758) = 8.566. P(Z >
8.566
1.96) = FZ(1.96)=.975
5.97 Portfolio consists of 10 shares of stock A and 8 shares of stock B.
a. Find the mean and variance of the portfolio value: W = 10X + 8Y with
correlation of .3.
W  a  x  b y = 10(10) + 8(12) = 196
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 102(16) + 82(9) + 2(10)(8)(.3)(4)(3) = 2,752
b. Option 1: Stock 1 with mean of 10, variance of 25, correlation of -.2.
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 102(25) + 82(9) + 2(10)(8)(-.2)(5)(3) = 2,596
Option 2: Stock 2 with mean of 10, variance of 9, correlation of .6.
= 102(25) + 82(9) + 2(10)(8)(.6)(5)(3) = 2,340
To reduce the variance of the porfolio, select Option 2
5.98 Portfolio consists of 10 shares of stock A and 8 shares of stock B
a. Find the mean and variance of the portfolio value: W = 10X + 8Y with
correlation of .3.
W  a  x  b y = 10(12) + 8(10) = 200
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 102(14) + 82(12) + 2(10)(8)(.5)(3.74166)(3.4641) = 3,204.919
b. Option 1: Stock 1 with mean of 12, variance of 25, correlation of -.2.
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 122(25) + 82(12) + 2(10)(8)(-.2)(5)(3.4641) = 3,813.744
Option 2: Stock 2 with mean of 10, variance of 9, correlation of .6.
= 102(9) + 82(12) + 2(10)(8)(.6)(3)(3.4641) = 2,665.66
To reduce the variance of the porfolio, select Option 2
145
146
5.99
Statistics for Business & Economics, 7th edition
W  a  x  b y = 1(800000) + 1(60000) = 140000
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 12(1000000) + 12(810000) + 2(1)(1)(.4)(1000)(900) = 2,530,000
W  2,530,000  1590.597372
Probability that the weight is between 138,000 and 141,000:
138, 000  140, 000
141, 000  140, 000
= –1.26 fz = .3962,
= .63 fz = .2357
1590.597372
1590.597372
.3962 + .2357 = .6319
5.100 a. W  a  x  b y = 1(40) + 1(35) = 75
 2W  a 2 2 X  b 2 2Y  2abCorr ( X , Y ) X  Y
= 12(100) + 12(144) + 2(1)(1)(.6)(10)(12) = 388
W  388  19.69772
Probability that all seats are filled:
100  75
= 1.27 Fz = .8980. 1 – .8980 = .1020
19.69772
b. Probability that between 75 and 90 seats will be filled:
90  75
= .76 .5 – Fz(.76) = .2764
19.69772
5.101
Mean and variance for stock prices:
Alcoa Inc. Reliant Energy, Inc. Sea Containers Ltd.
Mean
31.98286
13.07000
10.55286
Variance
27.41879
54.08367
60.97572
Covariances:
Reliant Energy, Inc.
Sea Containers Ltd.
Alcoa Inc. Reliant Energy, Inc.
22.1591
5.6791
-27.7501
Let the total value of the portfolio be represented by variable W.
W = (0.3333)(31.98286) + (0.3333)(13.07000) + (0.3333)(10.55286) = 18.54
 W2 = (0.33332)(27.41879) + (0.33332)(54.08367) + (0.33332)(60.97572) +
2[(0.3333)(0.3333)(22.1591) + (0.3333)(0.3333)(5.679) +
(0.3333)(0.3333)(-27.7501)] = 15.85
Chapter 5: Continuous Random Variables and Probability Distributions
147
We can confirm these results by finding the portfolio price for each year, shown
next, and then by finding the mean and variance of these prices.
Portfolio Price
20.9567
14.9733
17.4767
21.5867
21.2033
11.6367
21.9133
Descriptive Statistics: Portfolio Price
Variable
Portfolio Price
N
7
N*
0
Variable
Portfolio Price
Q3
21.59
Mean
18.54
SE Mean
1.50
StDev
3.98
Variance
15.85
Minimum
11.64
Q1
14.97
Maximum
21.91
The previous output confirms that W = 18.54 and  W2 = 15.85.
Assuming that the portfolio price is normally distributed, the narrowest interval
that contains 95% of the distribution of portfolio value is centered at the mean.
Therefore, it is W  z / 2 W . Using z / 2  1.96 and  W  3.98, the interval is
18.54  (1.96)(3.98) or (10.74, 26.34).
5.102
Mean and variance for stock prices:
Mean
Variance
TCF Financial
AB Volvo (ADR) Alcoa Inc.
Pentair Inc. Corporation
8.6143
31.9829
28.9543
25.1643
25.4171
27.4188
95.4157
20.4361
Covariances:
Alcoa Inc.
Pentair Inc.
TCF Financial
Corporation
AB Volvo (ADR)
6.5180
31.2128
Alcoa Inc.
-4.3594
-2.7947
Pentair Inc.
5.4712
20.6897
Median
20.96
148
Statistics for Business & Economics, 7th edition
Let the total value of the portfolio be represented by variable W.
W = (0.3333)(8.6143) + (0.1667)(31.9829) + (0.3333)(28.9543) +
(0.1667)(25.1643) = 22.05
 W2 = (0.33332)(25.4171) + (0.16672)(27.4188) + (0.33332)(95.4157) +
(0.16672)(20.4361) + 2[(0.3333)(0.1667)(6.5180) + (0.3333)(0.3333)(31.2128) +
(0.3333)(0.1667)(-4.3594) + (0.1667)(0.3333)(5.4712) +
(0.1667)(0.1667)(-2.7947) + (0.3333)(0.1667)(20.6897)] = 24.68
We can confirm these results by finding the portfolio price for each year, shown
next, and then by finding the mean and variance of these prices.
Portfolio Price
26.1833
24.6217
24.0983
27.7533
20.2700
14.3017
17.1033
Descriptive Statistics: Portfolio Price
Variable
Portfolio Price
N
7
N*
0
Variable
Portfolio Price
Q3
26.18
Mean
22.05
SE Mean
1.88
StDev
4.97
Variance
24.68
Minimum
14.30
Q1
17.10
Median
24.10
Maximum
27.75
The previous output confirms that W = 22.05 and  W2 = 24.68.
Assuming that the portfolio price is normally distributed, the narrowest interval
that contains 95% of the distribution of portfolio value is centered at the mean.
Therefore, it is W  z / 2 W . Using z / 2  1.96 and  W  4.97, the interval is
22.05  (1.96)( 4.97) or (12.31, 31.79).
5.103
Mean and variance for stock prices:
Mean
Variance
General
Sea
3M
Alcoa Intel
Potlatch
Motors
Containers
Company Inc.
Corporation Corporation Corporation
Ltd.
75.373 31.983
24.904
39.684
36.279
10.553
113.671 27.419
34.680
111.229
150.734
60.976
Chapter 5: Continuous Random Variables and Probability Distributions
149
Covariances:
3M
Alcoa
Intel
Potlatch
Company Inc.
Corporation Corporation
25.5503
14.0721 28.3612
81.3497 9.8462
2.9658
Alcoa Inc.
Intel Corporation
Potlatch Corporation
General Motors
Corporation
Sea Containers Ltd.
-28.2769 23.1876
-1.2885 5.6791
32.7968
17.3941
General Motors
Corporation
-74.5682
-2.1902
57.4105
Let the total value of the portfolio be represented by variable W. The mean and
variance for this portfolio, W and  W2 , can be found using the following
equations or by using technology.
k
k
k 1
i 1
i 1
i 1
W   ai  i ,  W2   ai2 i2  2
k
a a
j i 1
i
j
Cov( X i , X j )
Descriptive Statistics: Portfolio Price
Variable
Portfolio Price
N
7
N*
0
Variable
Portfolio Price
Q3
41.20
Mean
36.46
SE Mean
1.87
StDev
4.95
Variance
24.54
Minimum
28.28
Q1
33.91
Median
36.16
Maximum
43.58
As previously shown, W = 36.46 and  W2 = 24.54.
Assuming that the portfolio price is normally distributed, the narrowest interval
that contains 95% of the distribution of portfolio value is centered at the mean.
Therefore, it is W  z / 2 W . Using z / 2  1.96 and  W  4.95, the interval is
36.46  (1.96)( 4.95) or (26.76, 46.16).
5.104
Mean and variance for stock price growth:
3M
Alcoa
Intel
Potlatch
General Motors Sea Containers
Company Inc.
Corporation Corporation Corporation
Ltd.
Mean
0.001992 0.004389 -0.000082
0.007449
-0.014355
-0.146323
Variance 0.002704 0.005060
0.006727
0.006674
0.014518
0.176663
150
Statistics for Business & Economics, 7th edition
Covariances:
Alcoa Inc.
Intel Corporation
Potlatch Corporation
General Motors
Corporation
Sea Containers Ltd.
General
Motors
Corporation
3M
Intel
Potlatch
Company
Alcoa Inc.
Corporation
Corporation
0.00153782
0.00163165 0.00184360
0.00012217 0.00197600
0.00144736
-0.00005101 0.00103371
0.00075015 0.00706908
-0.00006588
-0.00131221
0.00246545
-0.00151704
0.01077420
Let the portfolio growth be represented by variable W. The mean and variance for this
portfolio, W and  W2 , can be found using the following equations or by using technology.
k
k
k 1
i 1
i 1
i 1
W   ai  i ,  W2   ai2 i2  2
k
a a
j i 1
i
j
Cov( X i , X j )
Descriptive Statistics: Portfolio Growth
Variable
Portfolio Growth
N
60
N*
0
Variable
Portfolio Growth
Median
-0.0062
Mean
-0.0245
SE Mean
0.0111
Q3
0.0303
Maximum
0.1212
StDev
0.0862
Variance
0.0074
Minimum
-0.4182
Q1
-0.0688
As previously shown, W = -0.0245 and  W2 = 0.0074.
5.105
Mean and variance for stock price growth:
General Motors International
Potlatch
Sea Containers Tata
Corporation
Business Machines Corporation Ltd.
Communications
Mean
-0.014355
0.004589
0.007449
-0.146323
0.022260
Variance
0.014518
0.002607
0.006674
0.176663
0.021645
Covariances:
General Motors
Corporation
General Motors
Corporation
International Business
Machines
Potlatch Corporation
Sea Containers Ltd.
Tata Communications
0.00061410
0.00246545
0.01077420
0.00108433
International
Potlatch
Business Machines Corporation
0.00097139
0.00256087
0.00149232
-0.00151704
0.00626864
Sea Containers
Ltd.
-0.00332721
Chapter 5: Continuous Random Variables and Probability Distributions
151
Let the portfolio growth be represented by variable W. The mean and variance for this
portfolio, W and  W2 , can be found using the following equations or by using
technology.
k
k
k 1
i 1
i 1
i 1
W   ai  i ,  W2   ai2 i2  2
k
a a
j i 1
i
j
Cov( X i , X j )
Descriptive Statistics: Portfolio Growth
Variable
Portfolio Growth
N
60
N*
0
Variable
Portfolio Growth
Median
-0.0048
Mean
-0.0253
SE Mean
0.0133
Q3
0.0461
Maximum
0.1378
StDev
0.1029
Variance
0.0106
Minimum
-0.4573
Q1
-0.0763
As previously shown, W = -0.0253 and  W2 = 0.0106.
For the second portfolio (40% General Motors, 30% International Business
Machines, and 30% Tata Communications), we get the following output:
Descriptive Statistics: Portfolio Growth
Variable
Portfolio Growth
N
60
N*
0
Mean
0.00231
Variable
Portfolio Growth
Q1
-0.05795
SE Mean
0.00929
Median
0.01323
StDev
0.07198
Q3
0.06036
Variance
0.00518
Minimum
-0.13973
Maximum
0.19402
For the second portfolio, as previously shown, W = -0.00231 and  W2 = 0.00518.
The second portfolio has a higher mean and a lower variance. Since risk is directly related
to variance, the second portfolio would be the better choice.
5.106
Mean and variance for stock price growth:
AB
Pentair
Reliant
Volvo
Inc.
Energy Inc.
Mean
0.019592 0.007641
0.019031
Variance 0.004805 0.006227
0.012686
TCF Financial
Corporation
-0.004087
0.004001
3M
Company
0.001992
0.002704
Restoration
Hardware Inc.
-0.013406
0.027618
152
Statistics for Business & Economics, 7th edition
Covariances:
Pentair Inc.
Reliant
Energy Inc.
TCF Financial
Corporation
3M Company
Restoration
Hardware Inc.
AB Volvo
0.00074848
Pentair Inc.
Reliant
Energy Inc.
TCF
Financial
Corporation
0.00072435
0.00228027
0.00105381
-0.00001514
0.00099279
-0.00021080
0.00087718
-0.00041228
0.00031032
0.00117969
0.00169410
0.00055922
3M
Company
-0.00041072 0.00204408
Let the portfolio growth be represented by variable W. The mean and variance for this
portfolio, W and  W2 , can be found using the following equations or by using technology.
k
k
k 1
i 1
i 1
i 1
W   ai  i ,  W2   ai2 i2  2
k
a a
j i 1
i
j
Cov( X i , X j )
Descriptive Statistics: Portfolio Growth
Variable
Portfolio Growth
N
60
N*
0
Variable
Portfolio Growth
Q1
-0.02762
Mean
0.00513
Median
0.00631
SE Mean
0.00612
StDev
0.04740
Q3
0.04184
Variance
0.00225
Minimum
-0.16714
Maximum
0.10438
As previously shown, W = 0.00513 and  W2 = 0.00225.
For the second portfolio (20% AB Volvo, 30% Pentair, 30% Reliant Energy, and
20% 3M Company), we get the following output:
Descriptive Statistics: Portfolio Growth
Variable
Portfolio Growth
N
60
N*
0
Variable
Portfolio Growth
Q1
-0.02121
Mean
0.01232
Median
0.01386
SE Mean
0.00680
StDev
0.05270
Q3
0.05357
Variance
0.00278
Minimum
-0.15522
Maximum
0.10539
For the second portfolio, as previously shown, W = 0.01232 and  W2 = 0.00278.
The second portfolio has a higher mean and a higher variance. Recall that risk is directly
related to variance. Since the second portfolio has a significantly larger mean and only a
slightly larger variance, it would be the better choice.