Chapter 6:
Continuous Random Variables and Probability
Distributions
6.1 P(1.4 < X < 1.8) = F(1.8) – F(1.4) = (.5)(1.8) – (.5)(1.4) = 0.20
6.2 P(1.0 < X < 1.9) = F(1.9) – F(1.0) = (.5)(1.9) – (.5)(1.0) = 0.45
6.3 P(X < 1.4) = F(1.4) = (.5)(1.4) = 0.7
6.4 P(X > 1.3) = F(1.3) = (.5)(2.0) – (.5)(1.3) = 0.35
6.5 a.
Probability Density Function: f(x)
1.5
f(x)
1.0
0.5
0.0
0
X
1
Chapter 6: 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
6.6 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, 6th 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
6.7 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
6.8 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
6.9 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.
6.10 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.
6.11 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.
Chapter 6: Continuous Random Variables and Probability Distributions
6.12
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
6.13
Y = 10,000 + 1.5 X = 10,000 + 1.5 (30,000) = $55,000
Y = |1.5| X = 1.5 (8,000) = $12,000
6.14
Y = 20 + X = 20 + 4 = $24 million
Bid = 1.1 Y =1.1(24) = $26.4 million,  = $1 million
6.15
Y = 60 + .2 X = 60 + 140 = $200
Y = |.2| X = .2 (130) = $26
6.16
Y = 6,000 + .08 X = 6,000 + 48,000 = $54,000
Y = |.08| X = .08(180,000) = $14,400
6.17
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
6.18
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
6.19
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
125
126
Statistics for Business & Economics, 6th edition
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
6.20
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
6.21
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
6.22
a. P(Z <
400  380
) = P(Z < .4) = .6554
50
360  380
b. P(Z >
) = P(Z > -.4) = FZ(.4) = .6554
50
c. The graph should show the property of symmetry – the area in the tails
equidistant from the mean will be equal.
300  380
400  380
d. P(
<Z<
) = P(-1.6 < Z < .4) = FZ(.4) – [150
50
FZ(1.6)] = .6554 - .0548 = .6006
Chapter 6: Continuous Random Variables and Probability Distributions
e. 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
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 =
100
100
.6826
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
6.23
a. P(Z >
6.24
a. P(Z >
6.25
a. P(Z >
6.26
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
127
128
Statistics for Business & Economics, 6th edition
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.
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).
6.27
a. P(
6.28
P(Z > 1.5) = 1 - Fz(1.5) = .0668
6.29
P(Z < -1.28) = .1, –1.28 =
6.30
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
6.31
a. P(Z >
6.32
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 6: 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
6.33
For Supplier A: P(Z <
6.34
a. P(Z > -1.28) = .9, -1.28 =
6.35
a. P(Z <
6.36
a. P(
6.37
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 =
80
.2922
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
6.38
Statistics for Business & Economics, 6th 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 >
6.39
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) =
15
.0918
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
6.40
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 than how many? Z =
-1.34.
X  1600
1.34 
X = 1573.741
19.5959
1,574 successes
Chapter 6: Continuous Random Variables and Probability Distributions
e.
Probability is .20 the number of successes is
X  1600
greater than? Z = .84. .84 
.
19.5959
X = 1616.46 1,616 successes
6.41
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
6.42
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, 6th edition
e. Probability is .09 the percentage of successes
X  .40
is greater than? Z = 1.34. 1.34 
. P=
.01225
41.642%
6.43
6.44
6.45
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) =
6
6
.9525 - .5 = .4525
Chapter 6: Continuous Random Variables and Probability Distributions
34  40
48  40
<Z<
) = P(-1 < Z < 1.33) = Fz (1.33) – [1 – FZ(1)]
6
6
= .9082 - .1587 = .7495
d. 39 - 41
c. P(
6.46
E[X] = (100)(.6) = 60,  =
P(Z <
6.47
6.48
6.49
6.50
(100)(.6)(.4) = 4.899
50  60
) = P(Z < -2.04) = 1 – FZ(2.04) = 1- .9793 = .0207
4.899
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
133
134
Statistics for Business & Economics, 6th edition
6.51  = 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
6.52
 = 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
6.53
 = 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
6.54
 = 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
6.55 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
Chapter 6: Continuous Random Variables and Probability Distributions
6.56
P(X > 18) = e  (18 /15) = .3012
6.57
P(X > 2) = e  (2)(.8) = .2019
6.58
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.
6.59
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
6.60
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
6.61 Find the mean and variance of the random variable: W = 5X – 4Y with
correlation = .5.
W  a  x  b y = 5(100) – 4(200) = -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
6.62 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
135
136
Statistics for Business & Economics, 6th edition
6.63 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
 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
6.64
6.65 Assume that costs are independent across years
 Z = 5  x = 5(200) = 1,000
Z =
5 x =
2
5(3,600) = 134.16
 Z = 1  2  3 = 50,000 + 72,000 + 40,000 = 162,000
6.66
Z =  1   2   3 =
2
2
2
(10, 000) 2  (12, 000) 2  (9, 000) 2 = 18,027.76
 Z = 1  2  3 = 20,000 + 25,000 + 15,000 = 60,000
6.67
Z =  1   2   3 =
2
2
2
(2, 000) 2  (5, 000) 2  (4, 000) 2 = 6,708.2
6.68 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
 Z = (16  x ) / 16 =  x = 28
6.69
Z = 16 x / 16 =
2
(2.4) 2 / 16 = 2.4 / 4 = .6
6.70 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
 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
Chapter 6: Continuous Random Variables and Probability Distributions
6.71
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
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
6.72 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
6.73
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 >
2000  2400
) = P(Z > -1.54) = FZ(1.54) = .9382
260
6.74 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 >
6.75
0  (5)
) = P(Z > .23) = 1 – FZ(.23) = 1 – .5910 = .4090
21.79449
a. P(X < 10) = (10/12) – (8/12) = 1/6
b. P(X > 12) = (20/12) – (12/12) = 8/12 = 2/3
137
138
Statistics for Business & Economics, 6th edition
c. E[] = (20/12 – 12/12) = 2(2/3) = 1.333
d. 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.
6.76 a.
Probability Density Function: f(x)
f(x)
0.033333
0.000000
30
35
40
45
50
55
60
65
70
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 6: Continuous Random Variables and Probability Distributions
6.77
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
6.78
6.79
6.80
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
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
6.81 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)
139
140
Statistics for Business & Economics, 6th 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
6.82 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).
d. P(X > 65) = .3085 (from part a)
Use the binomial formula: P(X = 2) = C24 (.3085)2 (.6915)2 = 0.2731
6.83
a. P(Z >
6.84
a. P(Z <
85  100
) = P(Z < -.5) = .3085
30
70  100
130  100
b. P(
<Z<
) = P(-1 < Z < 1) = 2 Fz (1) – 1 = .6826
30
30
Xi  100
c. P(Z > 1.645) = .05, 1.645 =
, Xi = 149.35
30
60  100
d. P(Z >
) = P(Z > -1.33) = FZ(1.33) = .9032
30
P(X  1) = 1 – P(X = 0) = 1 – (.0918)2 = .9916
Chapter 6: Continuous Random Variables and Probability Distributions
e. Use the binomial formula: P(X = 2) = C24 (.9082)2 (.0918)2 = 0.0417
f. 90 – 109
g. 130 - 149
6.85
b.
c.
d.
e.
f.
6.86
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 >
6.87

,  = 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
6.88
E[X] = 1000(.4) = 400, x = (1000)(.4)(.6) = 15.4919
500  400
P(Z <
) = P(Z < 6.45)  1.0000
15.4919
6.89
E[X] = 400(.6) = 240, x = (400)(.6)(.4) = 9.798
200  240
P(Z >
) = P(Z > -4.08)  1.0000
9.798
6.90
P(Z <
50  70
) = P(Z < -2.39) = 1 – FZ (2.39) = .0084
70
141
142
6.91
Statistics for Business & Economics, 6th edition
a. P(X = 6) =
e6 66
= .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
6.92
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
6.93
a. P(
120  132
150  132
<Z<
) = P(- 1 < Z < 1.5) = FZ (1.5) – [1 – FZ(1)]
12
12
= .7745
b. P(Z > .44) = .33, .44 =
Xi  132
, Xi = 137.28
12
120  132
) = 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
c. P(Z <
6.94
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
6.95 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
Chapter 6: Continuous Random Variables and Probability Distributions
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
6.96 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
6.97
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
6.98 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
143