ASBE 5e Solutions for Instructors
Applied Statistics in Business and
Economics 5th Edition Doane Seward
0077837304 9780077837303
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Chapter 6
Discrete Distributions
6.1
Example A is a probability distribution because the sum of P(x) is 1 (.8 + .2 = 1) and all
probabilities are nonnegative.
Examples B and C are not probability distributions because the sum of P(x) is .95 for B and
1.30 for C.
Learning Objective: 06-1
6.2
a. P(X = 75) = .3
b. P(X ≤ 75) = .2 + .4 + .3 = .9
c. P(X > 50) = .3 + .1 = .4
d. P(X < 100) = .2 + .4 + .3 = .9
e. (b) is a CDF because a CDF is defined by the inequality “≤”.
Learning Objective: 06-1
6.3
a. P(X ≥ 3) = .2 + .15 + .05 = .4
b. P(X ≤ 2) = .05 + .3 + .25 = .6
c. P(X > 4) = .05 + .3 + .25 + .2 = .8
d. P(X =1) = .3
e. (b) is a CDF because a CDF is defined by the inequality “≤”.
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ASBE 5e Solutions for Instructors
Learning Objective: 06-1
6.4
a. Based on the table below we see that: E(X) = 57.5, V(X) = 506.25,  = 22.5.
b. The distribution is skewed to the right.
x
P(x)
xP(x)
x−E(X)
P(x)[x−E(X)]2
25
50
75
100
Total
0.20
0.40
0.30
0.10
1.00
5.00
20.00
22.50
10.00
57.50
-32.50
-7.50
17.50
42.50
211.25
22.5
91.875
180.625
506.25
Learning Objective: 06-2
6.5
a. Based on the table below: E(X) = 2.25, V(X) = 1.6875,  = 1.299.
b. The distribution is skewed to the right.
x
P(x)
xP(x)
x−E(X)
P(x)[x−E(X)]2
0
1
2
3
4
5
Total
0.05
0.30
0.25
0.20
0.15
0.05
1.00
0.00
0.30
0.50
0.60
0.60
0.25
2.25
-2.25
-1.25
-0.25
0.75
1.75
2.75
0.25
0.47
0.02
0.11
0.46
0.38
1.6875
Learning Objective: 06-2
6.6
X is the amount of the “value” of a bottle. X can be either $215,000, the value of the
Lamborghini, or $0. To find the expected value multiply each value of X by the
associated probability. E(X) = ($215,000)(.00000884) + ($0)(.99999116) = 1.9006.
Learning Objective: 06-2
6.7
X is the amount of the claim. Assume a student who files a claim will claim the maximum
amount of $5000. The values of X are either $5000 or $0. Expected payout = E(X) =
$5000(.01) + ($0)(.999) = $50, so company adds $25 and charges $75.
Learning Objective: 06-2
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ASBE 5e Solutions for Instructors
6.8
X is the lottery ticket winnings. The values of X are either $28 million or $0. The expected
winning is E(X) = $28,000,000(.000000023) + $0(.999999977) = $0.644. Because the
lottery ticket costs more than the expected winnings ($1.00 > $0.644) the ticket is too
high to buy.
Learning Objective: 06-2
6.9
Let X equal the loss during a hurricane. The values of X are either $250 million, $950
million, or $0. Expected Loss = E(X ) = $250(.3) + $950(.3) + $0(.4) = $360 million.
Learning Objective: 06-2
6.10
Using a = 0000 and b = 9999, μ =(0000 + 9999)/2 = 4999.5.
 = [(9999 − 0 + 1)2 − 1] /12 = 2886.75
Learning Objective: 06-3
6.11
a. With a = 20 and b = 60,  = (20 + 60) / 2 = 40 and  = [(60 − 20 + 1) 2 − 1] /12 = 11.83
39 − 20 + 1
b. P(X  40) = 1− P(X ≤ 39) = 1−
= .5122.
60 − 20 + 1
29 − 20 + 1
P(X  30) = 1− P(X ≤ 29) = 1−
= .7561.
60 − 20 + 1
Learning Objective: 06-3
6.12
a. Answers will vary but the excel function is =RANDBETWEEN(1,2) and the mean and
standard deviation of the numbers generated should be close to  = 1.5 and  = .50
because
a + b 1+ 2
[(b − a) + 1]2 − 1
[(2 − 1) + 1]2 − 1
=
= 1.5 and
=
= .5
12
12
2
2
b. Answers will vary but the excel function is =RANDBETWEEN() and the mean and
standard deviation of the numbers generated should be close to  = 3.0 and  = 1.41.
c. Answers will vary but the excel function is =RANDBETWEEN() and the mean and
standard deviation of the numbers generated should be close to  = 49.5 and  = 28.87.
d. See parts a thru c.
e. See parts a thru c.
Learning Objective: 06-3
6.13
a. X = 0, 1, or 2
b. X = 4, 5, 6, or 7
c. X = 4, 5, 6, 7, 8, 9, or 10
Learning Objective: 06-4
6.14
a. P( X  7)
b. P( X < 4)
c. P( X  2)
Learning Objective: 06-4
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ASBE 5e Solutions for Instructors
6.15
a.  = (8)(.1) = 0.8,  = (8)(.1)(1 − .1) = 0.8485
b.  = (10)(.4) = 4,  = (10)(.4)(1 − .4) = 1.5492
c.  = (12)(.5) = 6,  = (12)(.5)(1 − .5) = 1.7321
Learning Objective: 06-4
6.16
a.  = (30)(.9) = 27,  = (30)(.9)(1 − .9) = 1.6432
b.  = (80)(.7) = 56,  = (80)(.7)(1 − .7) = 4.0988
c.  = (20)(.8) = 16,  = (20)(.8)(1 − .8) = 1.7889
Learning Objective: 06-4
9!
.95 (1 − .9)9−5 = .0074
5!(9 − 5)!
6!
.20 (1 − .2)6−0 = .2621
b. P( X = 0) =
0!(6 − 0)!
9!
.89 (1 − .8)9−9 = .1342
c. P( X = 9) =
9!(9 − 9)!
Learning Objective: 06-4
6.17
a. P( X = 5) =
6.18
a. P( X = 2) =
6.19
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative.
a. P(X ≤ 3) =BINOM.DIST(3,8,.2,1) = .9437
b. P(X > 7) = 1 – P(X ≤ 7) = 1-BINOM.DIST(7,10,.5,1) = .0547
c. P(X < 3) = P(X ≤ 2) =BINOM.DIST(2,6,.7.1) = .0705
Learning Objective: 06-4
6.20
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative.
a. P(X ≤ 10) =BINOM.DIST(10,14,.95,1) = .00417
b. P(X > 2) = 1 – P(X ≤ 2) = 1-BINOM.DIST(2,5,.45,1) = .4069
c. P(X ≤ 1) =BINOM.DIST(1,10,.15,1) = .5443
Learning Objective: 06-4
6.21
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative.
a. P(X > 10) = 1 – P(X ≤ 10) =1-BINOM.DIST(10,16,.8,1) = .9183
8!
.102 (1 − .10)8−2 = .1488
2!(8 − 2)!
10!
b. P( X = 1) =
.401 (1 − .40)10−1 = .0403
1!(10 − 1)!
12!
c. P( X = 3) =
.703 (1 − .70)12−3 = .0015
3!(12 − 3)!
Learning Objective: 06-4
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ASBE 5e Solutions for Instructors
b. P(X ≥ 4) = 1 – P(X ≤ 3) =1-BINOM.DIST(3,8,.4,1) = .4059
c. P(X ≤ 2) =BINOM.DIST(2,6,.2,1) = .9011
Learning Objective: 06-4
6.22
Use the Excel function =BINOM.DIST(x,n,π,1) where “1” stands for cumulative.
a. P(X < 4 ) = P(X ≤ 3) =BINOM.DIST(3,12,.10,1) = .9744.
b. P(X ≥ 3) = 1 – P(X ≤ 2) =1-BINOM.DIST(2,7,.40,1) = .5801.
c. P(X ≤ 9) =BINOM.DIST(9,14,.60,1) = .7207.
Learning Objective: 06-4
6.23
The number of customers out of the next 10 who pay with American Express is a binomial
random variable with n = 10 and π = .2.
a. P(X = 0) =BINOM.DIST(0,10,.2,0) = .1074.
b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-BINOM.DIST(1,10,.2,1) = .6242.
c. P(X < 3) = P(X ≤ 2) =BINOM.DIST(2,10,.2,1) = .6778
d. μ = nπ =(10)(.2) = 2
e. σ = (10)(.2)(1 − .2) = 1.2649
f. See below.
g. Skewed to the right.
Learning Objective: 06-4
6.24
The number of customers out of 12 who have an incorrect address in the database is a
binomial random variable with n = 12 and π = .05.
12!
a. P( X = 0) =
.050 (1 − .05)12−0 = .5404 or use Excel: =BINOM.DIST(0,12,.05,0).
0!(12 − 0)!
12!
b. P( X = 1) =
.051 (1 − .05)12−1 = .3413 or use Excel: =BINOM.DIST(1,12,.05,0).
1!(12 − 1)!
12!
c. P( X = 2) =
.052 (1 − .05)12−2 = .0988 or use Excel: =BINOM.DIST(2,12,.05,0).
2!(12 − 2)!
d. P( X  3) = P( X  2) = P( X = 0) + P( X = 1) + P( X = 2) = .9804. Either add parts a-c
above or use the Excel function =BINOM.DIST(2,12,.05,1) = .9804.
e. See below, skewed to the right.
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ASBE 5e Solutions for Instructors
Learning Objective: 06-5
6.25
The number of customers out of the next 5 who order a nonalcoholic beverage is a binomial
random variable with n = 5 and π = .38.
a. P(X =0) =BINOM.DIST(0,5,.38,0) = .0916
b. P(X ≥ 2) = 1–P(X ≤ 1) =1- BINOM.DIST(1,5,.38,1) = .6276
c. P(X < 4) = P(X ≤ 3) =BINOM.DIST(3,5,.38,1) = .9274
d. P(X =5) =BINOM.DIST(5,5,.38,0) = .0079
Learning Objective: 06-4
6.26
The number of car buyers out of a sample of 8 who use the Internet for research is a binomial
random variable with n = 8 and π = .6.
8!
a. P( X = 8) =
.608 (1 − .60)8−8 = .0168 or use Excel: =BINOM.DIST(8,8,.6,0).
8!(8 − 8)!
b. P( X  5) = 1 − P( X  4) =1-BINOM.DIST(4,8,.60,) = .5941.
c. P( X > 4) = 1 − P( X  4) =1-BINOM.DIST(4,8,.60,1) = .5941. Note that we could also
use the answer from part b because P( X  5) = P( X  4) .
d.  = n = (8)(.60) = 4.8 and  = 8(.6)(.4) = 1.3856 .
e. It is almost symmetric (slightly left-skewed).
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ASBE 5e Solutions for Instructors
Learning Objective: 06-4
6.27
The number of passengers out of a sample of 16 who check their bags is a binomial random
variable with n = 16 and π = .7.
16!
(.7)16 (1-.7)0 = .0033 or use Excel: =BINOM.DIST(16,16,.7,0).
a. P(X = 16) =
0!16!
b. P(X < 10) = P(X ≤ 9) =BINOM.DIST(9,16,.7,1) = .1753.
c. P(X ≥ 10) = 1− P(X ≤ 9) =1-BINOM.DIST(9,16,.7,1) = .8247.
Learning Objective: 06-4
6.28
The number of drivers stopped for speeding out of a sample of 12 who have an invalid
drivers license is a binomial random variable with n = 12 and π = .15.
12!
a. P( X = 0) =
.150 (1 − .15)12−0 = .1422 or use Excel: =BINOM.DIST(0,12,.15,0).
0!(12 − 0)!
12!
b. P( X = 1) =
.151 (1 − .15)12−1 = .3012 or use Excel: =BINOM.DIST(1,12,.15,0).
1!(12 − 1)!
c. P( X  2) = 1 − P( X  1) =1-BINOM.DIST(1,12,.15,1) = .5565.
Learning Objective: 06-4
6.29
a.  = 1,  =  = 1.0 and  =
b.  = 2,  =  = 2.0 and  =
c.  = 4,  =  = 4.0 and  =
Learning Objective: 06-5
6.30
LO6
 = 1.
 = 1.414.
 = 2.0
a.  = 9,  =  = 9.0 and  =  = 3.
b.  = 12,  =  = 12.0 and  =  = 3.464.
c.  = 7,  =  = 7.0 and  =  = 2.646.
Learning Objective: 06-5
(4.0)6 (e) −4.0
= .1042 . Can also use Excel:
6.31 a.
 = 4, P( X = 6) =
6!
=POISSON.DIST(6,4,0).
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ASBE 5e Solutions for Instructors
(12.0)10 (e) −12.0
= .1048 . Can also use Excel:
10!
=POISSON.DIST(10,12,0).
(7.0) 4 (e) −7.0
= .0912 . Can also use Excel: =POISSON.DIST(4,7,0).
c.  = 7, P( X = 4) =
4!
Learning Objective: 06-5
b.  = 12, P( X = 10) =
6.32
(.1) 2 (e) −.1
= .0045 . Can also use Excel: =POISSON.DIST(2,.1,0).
a.  = 0.1, P( X = 2) =
2!
(2.2)1 (e) −2.2
= .2438 Can also use Excel: =POISSON.DIST(1,1.1,0).
b.  = 2.2, P( X = 1) =
1!
(1.6)3 ( e) −1.6
c.  = 1.6, P( X = 3) =
= .1378 Can also use Excel: =POISSON.DIST(3,1.6,0).
3!
Learning Objective: 06-5
6.33
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative.
a.  = 4.3, P(X  3) =POISSON.DIST(3,4.3,1) = .3772.
b.  = 5.2, P(X > 7) = 1 – P(X ≤ 7) =1-POISSON.DIST(7,5.2,1) =.1551.
c.  = 2.7, P(X < 3) = P(X ≤ 2) =POISSON.DIST(2,2.7,1) = .4936.
Learning Objective: 06-5
6.34
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative.
a.  = 11.0, P(X  10) =POISSON.DIST(10,11,1) = .4599.
b.  = 5.2, P(X > 3) = 1 − P(X ≤ 3) =1-POISSON.DIST(3,5.2,1) = .7619.
c.  = 3.7, P(X < 2) = P(X ≤ 1) =POISSON.DIST(1,3.7,1) = .1162.
Learning Objective: 06-5
6.35
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative.
a.  = 8.0, P(X > 10) = 1 − P(X  10) =1-POISSON.DIST(10,8,1) = .1841.
b.  = 4.0, P(X ≤ 5) =POISSON.DIST(5,4,1) = .7851.
c.  = 5.0, P(X ≥ 2) = 1 − P(X ≤ 1) =1-POISSON.DIST(1,5,1) = .9596.
Learning Objective: 06-5
6.36
Use the Excel function =POISSON.DIST(x,λ,1) where “1” stands for cumulative.
a.  = 5.8, P(X < 4) = P(X ≤ 3) =POISSON.DIST(3,5.8,1) =.1700.
b.  = 4.8, P(X ≥ 3) = 1 – P(X ≤ 2) =1-POISSON.DIST (2,4.8,1) = .8574.
c.  = 7.0, P(X ≤ 9) =POISSON.DIST (9,7,1) =.8305.
Learning Objective: 06-5
6.37
The number of problems per vehicle is a Poisson random variable with λ = 1.7.
a. P(X ≥ 1) = 1 – P(X = 0) =1-POISSON.DIST(0,1.7,0) = .8173.
b. P(X = 0) =POISSON.DIST(0,1.7,0) = .1827.
c. P(X > 3) = 1− P(X ≤ 3) =1-POISSON.DIST(3,1.7,1) = .0932
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ASBE 5e Solutions for Instructors
d. Skewed right.
Poisson distribution (µ = 1.7)
0.35
0.30
P(X)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
10
11 12
X
Learning Objective: 06-5
6.38
a. Cancellations are independent, come one at a time. We also know that cancellations are
discrete and that we are given a mean rate per unit of time.
(1.5)0 ( e) −1.5
b. P( X = 0) =
= .2231 Can also use Excel: =POISSON.DIST(0,1.5,0).
0!
(1.5)1 ( e) −1.5
c. P( X = 1) =
= .3347 Can also use Excel: =POISSON.DIST(1,1.5,0).
1!
d. P(X > 2) =1 − P(X ≤ 2) =1-POISSON.DIST(2,1.5,1) = .1912.
e. P(X  5) =1 − P(X ≤ 4) =1-POISSON.DIST(4,1.5,1) = .0186.
Learning Objective: 06-5
6.39
a. The number of add-ons is discrete and we’ll assume they “arrive” one at a time and are
independent.
b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-POISSON.DIST(1,1.4,1) = .4082.
c. P(X = 0) = POISSON.DIST(0,1.0,0) = .2466.
d. Skewed right.
Poisson distribution (µ = 1.4)
0.40
0.35
0.30
P(X)
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
X
Learning Objective: 06-6
6.40
a. Not independent events, the warm room leads to yawns from all.
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9
10
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ASBE 5e Solutions for Instructors
b. Answers will vary. Possible example: The occurrence of a contagious illness in a
particular geographic area. The fact the illness is contagious suggests the “arrivals” are
not independent.
Learning Objective: 06-5
6.41
Let  = n = (500)(.003) = 1.5
a. P(X ≥ 2) = 1 – P(X ≤ 1) =1- POISSON.DIST(1,1.5,1) = .4422.
b. P(X < 4) = P(X ≤ 3) = .9344.
c. Yes, based on our rule of thumb n  20 and   .05.
Learning Objective: 06-6
6.42
Let  = n = (100000)(.000002) = 2
a. P(X ≥ 1) = 1 − P(X=0) =1- POISSON.DIST(0,2,0) = .8647.
b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-POISSON.DIST(1,2,1) = .5940.
c. Yes, based on our rule of thumb n  20 and   .05.
Learning Objective: 06-6
6.43
a. μ = n = (200)(.03) = 6 letters.
b.  = (200)(.03)(1 − .03) = 2.412.
c. Using the Poisson approximation to the binomial:
P(X ≥ 10) = 1 – P(X ≤ 9) =1-POISSON.DIST(9,6,1) = .0839. If using Excel to calculate
the exact binomial: P(X ≥ 10) =1-BINOM.DIST(9,200,.03,1) = .0808.
d. Using the Poisson approximation to the binomial:
P(X ≤ 4) =POISSON.DIST(4,6,1) = .2851. If using Excel to calculate the exact binomial:
P(X ≤ 4) =BINOM.DIST(4,200,.03,1) = .2810.
e. Yes, based on our rule of thumb n  20 and   .05
Learning Objective: 06-6
6.44
Let  = n = (100)(.01) = 1. P(X ≥ 2) = 1 – P(X ≤ 1) =1-POISSON.DIST(1,1,1) = .2642.
Learning Objective: 06-6
6.45
a. Expected Value = μ = n = ()() = 
b. Poisson approximation: P(X = 0) = POISSON.DIST(0,2.3,0) = .1003
P(X > 2) =1 − P(X ≤ 2) =1-POISSON.DIST(2,2.3,1) = .4040.
c. The approximation is fairly accurate because n  20 and   .05.
Learning Objective: 06-6
6.46
a. i. X = 0, 1, 2, or 3
ii. X = 0, 1, 2, or 3
iii. X = 0, 1, 2, 3, or 4
iv. X = 0, 1, 2, 3, 4, 5, 6, or 7
C
C
C
C
b. i. P(X = 3) = s x N −s n− x = 4 3 10−4 3−3 = .0333
N Cn
10 C3
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ASBE 5e Solutions for Instructors
C x N −s Cn− x 3 C2 20−3 C5−2
=
= .1316
N Cn
20 C5
C
C
C
C
iii. P(X = 1) = s x N −s n− x = 9 1 36−9 4−1 = .4469
N Cn
36 C4
C
C
C
C
iv. P(X = 3) = s x N −s n− x = 10 3 50−10 7−3 = .1098
N Cn
50 C7
c. i. =HYPGEOM.DIST(3,3,4,10,0) = .0333
ii. =HYPGEOM.DIST(2,5,3,20,0) = .1316
iii. =HYPGEOM.DIST(1,4,9,36,0) = .4469
iv. =HYPGEOM.DIST(3,7,10,50,0) = .1098
Learning Objective: 06-7
ii. P(X = 2) =
6.47
s
a.
b. The distribution is symmetric with a small range (2 to 4).
Learning Objective: 06-7
6.48
a. Let X = number of essay questions the student receives. Using Excel with N = 60, n = 10,
and s = 6:
P(X = 0) =HYPGEOM.DIST(0,10,6,60,0) = .3174
P(X = 1) =HYPGEOM.DIST(1,10,6,60,0) = .4232
P(X = 2) =HYPGEOM.DIST(2,10,6,60,0) = .2070
P(X = 3) =HYPGEOM.DIST(3,10,6,60,0) = .0470
P(X = 4) =HYPGEOM.DIST(4,10,6,60,0) = .0051
P(X = 5) =HYPGEOM.DIST(5,10,6,60,0) = .0000
P(X = 6) =HYPGEOM.DIST(6,10,6,60,0) = .0000 and all of these added together = 1.
b. P(X = 0) = =HYPGEOM.DIST(0,10,6,60,0) = .3174
c. P(X  1) = 1 − P(X = 0) =1-HYPGEOM.DIST(0,10,6,60,0) = .6826.
d. P(X  2) = 1 − P(X  1) =1-HYPGEOM.DIST(1,10,6,60,1) = .2954.
e. Skewed right.
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ASBE 5e Solutions for Instructors
Learning Objective: 06-7
6.49
a. Let X = number of incorrect vouchers in sample. Use Excel with N = 50, n = 5, and s =
20.
b. P(X = 0) =HYPGEOM.DIST(0,5,20,50,0) = .0673
c. P(X = 1) =HYPGEOM.DIST(1,5,20,50,0) = .2587
d. P(X  3) = 1 − P(X  2) =1-HYPGEOM.DIST(2,5,20,50,1) = .3100
e. Fairly symmetric or slightly skewed right.
Learning Objective: 06-7
6.50
a. Let X = number of HIV specimens in sample. Using Excel with N = 40, n = 5 and s = 8:
P(X = 0) =HYPGEOM.DIST(0,5,8,40,0) = .3060
P(X = 1) =HYPGEOM.DIST(1, 5,8,40,0) = .4372
P(X = 2) =HYPGEOM.DIST(2, 5,8,40,0) = .2111
P(X = 3) =HYPGEOM.DIST(3, 5,8,40,0) = .0422
P(X = 4) =HYPGEOM.DIST(4, 5,8,40,0) = .0034
P(X = 5) =HYPGEOM.DIST(5, 5,8,40,0) = .0000
b. P(X = 0) =HYPGEOM.DIST(0,5,8,40,0) = .3060
c. P(X < 3) = P(X  2) =HYPGEOM.DIST(2,5,8,40,1) =.9543.
d. P(X  2) = 1 − P(X  1) =1-HYPGEOM.DIST(1, 5,8,40,1) = .2568.
e. Skewed right.
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ASBE 5e Solutions for Instructors
Learning Objective: 06-7
6.51
a. n/N = 3/100 = .03 < .05 therefore it is safe to use binomial approximation with π = s/N =
40/100 = .4.
Binomial approximation: P(X = 3) =BINOM.DIST(3,3,.4,0) = .0640.
Exact probability: P(X = 3) =HYPGEOM.DIST(3,3,40,100,0) = .0611.
b. n/N = 10/200 = .05 therefore it is probably safe to use binomial approximation with π =
s/N = 60/200 = .3.
Binomial approximation: P(X = 2) =BINOM.DIST(2,10,.3,0) = .2335.
Exact probability: =HYPGEOM.DIST(2,10,60,200,0) = .2354.
c. n/N = 12/160 = .075 > .05 therefore it isn’t safe to use binomial approximation. If one
were to use the binomial approximate π = s/N = 16/160 = .10.
Binomial approximation: P(X = 1) =BINOM.DIST(1,12,.1,0) = .3766.
Exact probability: =HYPGEOM.DIST(1,12,16,160,0) = .3885.
d. n/N = 7/500 = .014 < .05, therefore it is safe to use binomial approximation with π = s/N
= 350/100 = .7.
Binomial approximation: P(X = 5) =BINOM.DIST(5,7,.7,0) = .3177.
Exact probability: =HYPGEOM.DIST(5,7,350,500,0) = .3198.
Learning Objective: 06-8
6.52
a. X = the number of vouchers with errors. X has a hypergeometric distribution with N =
200, s = 20, and n = 5. Because n/N < .05 it is safe to use the binomial approximation
with π = s/N = 20/200 = .1.
b. P(X = 0) =BINOMDIST(0,5,.1,0) = .5905
c. P(X  2) = 1–P(X  1) =1-BINOMDIST(1,5,.1,1) = .0815.
Learning Objective: 06-8
6.53
a. X = the number of firearms backgrounds checks that show a felony conviction. X has a
hypergeometric distribution with N = 500, s = 50, and n = 10. Because n/N = 10/500 =
.02 < .05 it is safe to use the binomial with  = s/N= 50/500 = .1.
b. P(X=0) =BINOM.DIST(0,10,.1,0) = .3487.
c. P(X  2) = 1 – P(X ≤ 1) =1-BINOM.DIST(1,10,.1,1) = .2639.
d. P(X < 4) = P(X ≤ 3) =BINOM.DIST(3,10,.1,1) = .9872.
Learning Objective: 06-8
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ASBE 5e Solutions for Instructors
6.54
a. X = the number of vehicles that comply with California emissions law. X has a
hypergeometric distribution with N = 400, s = 320, and n = 6. Because n/N = 6/400 =
.015 < 0.05 it is safe to use the binomial approximation with π = s/N = 320/400 = .8.
b. P(X = 6) =BINOM.DIST(6,6,.8,0) = .2621.
c. P(X ≥ 4) = 1 − P(X ≤ 3) =1-BINOMDIST(3,6,.8,1) = .9011.
Learning Objective: 06-8
6.55
a. P(X = 5) = (.5)(1−.5)5-1 = .0313.
b. P(X = 3) = (.25)(1−.25)3-1 = .1406.
c. P(X = 4) = (.6)(1−.6)4-1 = .0384.
Learning Objective: 06-9
6.56
a. Geometric mean is 1/ = 1/(.20) = 5
b. Using the geometric CDF, P(X ≤ 10) = 1 − (1 − )x = 1 − (1 − .20)10 = .8926.
Learning Objective: 06-9
6.57
a. Geometric mean is 1/ = 1/(.50) = 2
b. Using the geometric CDF, P(X > 10) = 1 − [1 − (1−.50)10 ]= .5010 = .00098
Learning Objective: 06-9
6.58
a.  = 79.62.54 = 202.184 cm
b.  = 3.242.54 = 8.2296 cm
c. Rule 1 for the mean and Rule 2 for the standard deviation.
Learning Objective: 06-10
6.59
a. Applying Rule 3, we add the means for each month to get  = 9500+7400 + 8600 =
$25,500. Applying Rule 4, we add the variances for each month and then take the square
root of this sum to find the standard deviation for the quarter: 2 = 1250+1425+1610 =
4285. σ = (4285).5 = 65.4599.
b. Rule 4 assumes that the sales for each month are independent of each other. This may
not be valid, given that a prior month’s sales usually influence the next month’s sales.
Learning Objective: 06-10
6.60
 = (5/9)(37.1)–17.78 = 2.83° C,  = (5/9)(10.3) = 5.72° C
Learning Objective: 06-10
6.61
a. Let Y = Bob’s point total. . Y = (5)(80) = 400,  Y = (25)(5) = 11.18
b. No, given that the standard deviation is 11.18, 450 is more than three standard deviations
from the mean.
Learning Objective: 06-10
6.62
The probability of a payout is 1 − .99863 = .00137. The expected payout is
(.00137)(1,000,000) = $1,370 dollars. To break even, the company would charge $1,370.
Learning Objective: 06-2
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ASBE 5e Solutions for Instructors
6.63
The expected loss is E(X) = ($250)(.02) + ($0)(.98) = $5. This exceeds the $4 cost of
insurance (assuming you would lose the entire cost of the PDA). Statistically, it is worth
it to insure to obtain “worry-free” shipping, despite the small likelihood of a loss.
Learning Objective: 06-2
6.64
a. If uniform,  = (1 + 5)/2 =3 and  = [(5 − 1 + 1) 2 − 1] /12 = 1.414 .
b. Answers will vary.
c. Answers will vary.
d. =RANDBETWEEN(1,5)
Learning Objective: 06-3
6.65
a.
b.
c.
d.
e.
6.66
This is a discrete binomial probability problem with "success" as "order bread" and "failure"
as "do not order bread". The probability of success, π, is .14 and the number of trials, n,
is 10.
a. P(X > 5) = 1–P(X ≤ 5). Using Excel: =1-BINOM.DIST(5,10,.14,1) = 1– .99905 = .00095.
b. P(X ≤ 2). Using Excel: =BINOM.DIST(2,10,.14,1) = .8455.
c. P(X = 0). Using Excel: =BINOM.DIST(0,10,.14,0) = .2213.
d. From MegaStat we obtain the probability distribution and probability histogram. The
distribution is skewed to the right. Or we could recognize that because π < .5 the
distribution is skewed right.
 = .80 (answers will vary).
 = .300 (answers will vary).
 = .50 (answers will vary).
 = .80 (answers will vary).
Outcomes of one trial might influence the next. For example, if I fail to make a free
throw because I shot the ball “long”, I will adjust my next shot to be a little “shorter,”
hence, violating the independence rule.
Learning Objective: 06-4
107
ASBE 5e Solutions for Instructors
Binomial
distribution
X
0
1
2
3
4
5
6
7
8
9
10
P(X)
0.22130
0.36026
0.26391
0.11457
0.03264
0.00638
0.00086
0.00008
0.00000
0.00000
0.00000
n
p
Binomial distribution (n = 10, p = 0.14)
0.40
cumulative
probability
0.22130
0.58156
0.84547
0.96004
0.99267
0.99905
0.99991
0.99999
1.00000
1.00000
1.00000
0.35
0.30
0.25
P(X)
10
0.14
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
10
X
Learning Objective: 06-4
6.67
Define X to be the number of vehicles that fail the emissions test. X is binomial with n =10
and π = .1.
a. P(X = 0) = .5905. Using Excel: =BINOM.DIST(0,5,.1,0) .
b. P(X = 1) = .3281. Using Excel: =BINOM.DIST(1,5,.1,0).
c. Strongly skewed to the right.
Learning Objective: 06-4
6.68
This is a discrete binomial probability problem with "success" as "can transact business in a
foreign language" and "failure" as "cannot transact business in a foreign language".
Probability of success, π, is .20 and the number of trials, n, is 10.
a. P(X = 0) = BINOM.DIST(0,10,.20,0) = .1074.
b. P(X ≥ 2) = 1 – P(X ≤ 1). Using Excel: =1-BINOM.DIST(1,10,.20,1) = .6242.
c. P(X = 10) =BINOM.DIST(10,10,.20,0) = .00000
Learning Objective: 06-4
6.69
This is a discrete binomial probability problem with "success" as "crispy" and "failure" as
"original". Probability of success, π, is .50 and the number of trials, n = 4.
a. P(X = 0) =BINOM.DIST(0,4,.5,0) = .0625
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ASBE 5e Solutions for Instructors
b. P(X ≥ 2) = 1 – P(X ≤ 1) =1-BINOM.DIST(1,4,.5,1) = .6875
c. P(X ≤ 2) =BINOM.DIST(2,4,.5,1) = .6875
d. Symmetric.
Learning Objective: 06-4
6.70
This is a discrete binomial probability problem with "success" as "order light beer" and
"failure" as "do not order light beer". The probability of success, π, is .40
and the
number of trials, n = 8.
a. P(X = 0). Using Excel: =BINOM.DIST(0,8,.40,0) = .0168.
b. P(X = 1). Using Excel: =BINOM.DIST(1,8,.40,0) = .0896.
c. P(X = 2). Using Excel: =BINOM.DIST(2,8,.40,0) = .2090.
d. P(X ≤ 2). Using Excel: =BINOM.DIST(2,8,.40,1) = .3154.
e. Slightly skewed right.
Learning Objective: 06-4
6.71
a.
b.
c.
d.
P(X = 3) =BINOM.DIST(3,20,0.3,0) = .0716.
P(X = 7) =BINOM.DIST(7,50,0.1,0) = .1076.
P(X ≤ 6) =BINOM.DIST(6,80,0.05,1) = .8947.
P(X ≥ 30) = 1 – P(X ≤ 29) =1-BINOM.DIST(29,120,0.2,1) = .1067.
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ASBE 5e Solutions for Instructors
Learning Objective: 06-4
6.72
a. This is a discrete binomial probability problem where "success" can be "got a sentence
correct" and failure can be "got a sentence wrong". There are 20 sentences so we can say
there are 20 trials and the probability of getting one correct if randomly guessing is .50.
P(X ≥ 14) = 1 – P(X ≤ 13). Using Excel: =1-BINOM.DIST(13,20,.50,1) = .0577. The
probability of passing is about 6%; not very likely to pass when you guess.
b. Using the same approach as in part a, we can plug in various values of a "passing score"
and we can see that P(X ≥ 15) = .0207 therefore a score of 15 would be needed in order to
get a 5% or less probability that someone can pass by guessing.
Learning Objective: 06-4
6.73
Define X to be the number of defaulted student loans out of a sample of 10. X is binomial
with n =10 and π = .07.
a. P(X = 0) =BINOM.DIST(0,10,.07,0) = .4840.
b. P(X ≥ 3) = 1 – P(X ≤ 2) =1-BINOM.DIST(2,10,.07,1) = .0283.
c. For this binomial,  = n = (10)(.07) = 0.7 defaults.
Learning Objective: 06-4
6.74
This is a discrete binomial probability problem where "success" is "something in the pocket"
and failure is "nothing in the pocket". The probability of success, π, is .08 and the
number of trials, n, is 14. Using Excel: P( X = 0) =BINOM.DIST(0,14,0.08,0) =
0.3112.
Learning Objective: 06-4
6.75
Define X to be the number of movie commercials advertising ‘R” rated movies. X is binomial
with n = 16 and  = .8
a. P(X ≥ 10) = 1 – P(X ≤ 9) =1-BINOM.DIST(9,16,.8,1) = .9733.
b. P(X < 8) = P(X ≤ 7) =BINOM.DIST(7,16,.8,1) = .0015.
Learning Objective: 06-5
6.76
a. P(X = 7) =POISSON.DIST(7,10,0) = .0901.
b. P(X = 3) =POISSON.DIST (3,10,0) = .0076.
c. P(X < 5) = P(X ≤ 4) =POISSON.DIST (4,10,1) = .0293.
d. P(X ≥ 11) = 1 – P(X ≤ 10) =1-POISSON.DIST (10,10,1) = .4170.
Learning Objective: 06-5
6.77
Let X = the number of no shows.
a. If n = 10 and  = .10, then P(X = 0) =BINOM.DIST(0,10,.1,0) = .3487.
b. If n = 11 and  = .10, then P(X ≥ 1) = 1 – P(X=0) =1-BINOM.DIST(0,11,.1,0) = 1 –
.3138 = .6862.
c. If they sell 11 seats, there is no way that more than 1 will be bumped. P(X > 1) = 0.
d. Let X = the number who do show up and set  = .90. We want P(X  10)  .95 so we use
Excel’s function = 1−BINOM.DIST(9,n,.9,TRUE) for various values of n. It turns out
that n = 13 will suffice because P(X  13) = .9658 > .95.
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ASBE 5e Solutions for Instructors
n
11
12
13
P(X  9)
0.30264
0.11087
0.03416
P(X  10)
0.69736
0.88913
0.96584
Learning Objective: 06-4
6.78
a. Let X be the number that are not working. As long as no more than 2 are not working,
he will have enough. P(X ≤ 2) =BINOM.DIST(2,10,0.2,1) = .6778.
b. Let X be the number that are working and set  = .8. We want P(X  8)  .95 so we use
Excel’s function =1−BINOM.DIST(7,n,0.8,TRUE) for various values of n. It turns out
that n = 13 will suffice.
n
10
11
12
13
P(X ≤ 7)
0.32220
0.16114
0.07256
0.03004
P(X  8)
0.67780
0.83886
0.92744
0.96996
Learning Objective: 06-4
6.79
a. Because calls to a fire station within a minute are most likely all about the same fire, the
calls are not independent.
b. Answers will vary.
Learning Objective: 06-5
6.80
a. This is a binomial probability with "success" as "catch fraudulent e-mail" and "failure" as
"do not catch fraudulent e-mail". The probability of success, π, is .20 and the number of
trials, n, is 16. The expected number that would be caught by such a filter is nπ = .20×16
= 3.2.
16!
b. P( X = 0) =
.200 (1 − .20)16−0 = .02815. Using Excel: =BINOM.DIST(0,16,.2,0).
0!(16 − 0)!
Learning Objective: 06-5
6.81
a. P(X = 5) =POISSON.DIST(5, 2.8, 0) = .0872.
b. P(X ≤ 5) =POISSON.DIST(5, 2.8, 1) = .9349.
c.  = ()() =  arrivals/5 minute interval
d. Independent time intervals
Learning Objective: 06-5
111
ASBE 5e Solutions for Instructors
6.82
a. This is a Poisson probability because broken bats are rare and independent events. Given
that λ = 1.0:
(10 )( e −1 )
= .3679. Using Excel: =POISSON.DIST(0,1,0).
P ( x = 0) =
0!
b. P(X ≥ 2) = 1 − P(X ≤ 1) = 1 − [P(X = 0) + P(X = 1)] = 1–.7358 = .2642.
P ( x = 0) =
(10 )( e −1 )
= .3679
0!
P( x = 1) =
(11 )( e −1 )
= .3679
1!
Using Excel: =1-POISSON.DIST(1,1,1) = .2642.
There is about a 26% chance of having at least 2 broken bats.
Learning Objective: 06-5
6.83
X is a Poisson random variable with λ = .3.
a. P(X = 0) =POISSON.DIST(0,.3,0) = .7408
b. P(X ≥ 2) =1-POISSON.DIST(1,.3,1) = .0369
Learning Objective: 06-5
6.84
a.
b.
c.
d.
Near collisions are random and independent events that occur one at a time.
P(X ≥ 1) = 1 – P(X = 0). Using Excel: =1-POISSON.DIST(0,1.2,0) = .6988.
P(X > 3) = 1 − P(X ≤ 3). Using Excel: =1-POISSON.DIST(3,1.2,1) = .0338.
See below.
Learning Objective: 06-5
112
ASBE 5e Solutions for Instructors
6.85
a. Assume that cancellations are independent of each other, occur randomly and one at a
time.
b. P(X = 0) =POISSON.DIST(0,1.5,0) = .2231.
c. P(X = 1) =POISSON.DIST(1,1.5,0) = .3347.
d. P(X > 2) = 1 – P(X ≤ 2) =1-POISSON.DIST(2,1.5,1) = .1912.
e. P(X ≥ 5) = 1 – P(X ≤ 4) =1-POISSON.DIST(4,1.5,1) = .0186.
Learning Objective: 06-5
6.86
X is a Poisson random variable with λ = 3.8/hour.
a. P(X = 0) =POISSON.DIST(0,3.8,0) = .0224.
b. P(X < 4) = P(X ≤ 3) =POISSON.DIST(3,3.8,1) = .4735.
c. P(X > 5) = 1 – P(X ≤ 5) =1-POISSON.DIST(5,3.8,1) = .1844.
Learning Objective: 06-5
6.87
a. We assume that paint defects are independent events, distributed randomly over the
surface and occur one at a time. For this problem, we would use a mean of  = 2.4
defects per 3 square meter area.
b. P(X = 0) =POISSON.DIST(0,2.4,0) = .0907.
c. P(X = 1) =POISSON.DIST(1,2.4,0) = .2177.
d. P(X ≤ 1) =POISSON.DIST(1,2.4,1) = .3084.
Learning Objective: 06-5
6.88
X = number of claimed dependents that are ineligible for insurance. X is a binomial random
variable with n = 7 and π = .02.
a. P(X =0) =BINOM.DIST(0,7,.02,0) = .8681.
b. P(X ≥ 1) = 1− P(X = 0) =1-BINOM.DIST(0,7,.02,0) = .1319.
Learning Objective: 06-4
6.89
Assume the arrival of a rogue wave follows a Poisson distribution. The ship is out for five
days so the average number of rogue waves per voyage will be:
.0377 waves 24hours 5days


=
= 4.524 per 5-day voyage.
hour
day
voyage
P(X ≥ 1) = 1 – P(X = 0) =1-POISSON.DIST(0,4.524,0) = .9892.
Learning Objective: 06-5
6.90
a. Earthquakes are random and independent events. No one can predict when they will
occur and we assume they occur one at a time. We are also given a mean rate of discrete
events per unit of time.
b. P(X < 3) = P(X ≤ 2). Using Excel: =POISSON.DIST(2,1.2,1) = .8795.
c. P(X > 5) = 1 − P(X ≤ 5). Using Excel: =1-POISSON.DIST(5,1.2,1) = .0015.
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ASBE 5e Solutions for Instructors
Learning Objective: 06-5
6.91
a. Crashes are unrelated events, can’t predict them, so they do happen randomly. We
assume crashes occur independently and one at a time.
b. P(X ≥ 1) = 1 – P(X = 0) =1-POISSON.DIST(0,2,0) = .8647.
c. P(X < 5) = P(X ≤ 4) =POISSON.DIST(4,2,1) = .9474.
d. Skewed to the right.
Learning Objective: 06-5
*6.92
Binomial n = 2500,  = .001 or Poisson with  = .001(2500) = 2.5 leaks per 2500 meters.
Using the Poisson approximation to the binomial:
a. P(X = 0). Using Excel: =POISSON.DIST(0,2.5,0) = .0821
b. P(X ≥ 3) = 1 – P(X ≤ 2). Using Excel: =1–POISSON.DIST(2,2.5,1) = .4562.
c. µ = 2.5 leaks per 2500 meters.
Learning Objective: 06-6
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ASBE 5e Solutions for Instructors
*6.93
a. Binomial n = 200,  = .02. Define X to be the number of twin births in 200 deliveries.
E(X) = (200)(.02) = 4.
b. Using the Poisson approximation: P(X = 0) =POISSON.DIST(0,4,0) = .0183. Using the
binomial function: P(X = 0) =BINOM.DIST(0,200,.02,0) = .0176.
c. Using the Poisson approximation: P(X = 1) =POISSON.DIST(1,4,0) = .0733. Using the
binomial function: P(X = 1) =BINOM.DIST(1,200,.02,0) = .0718.
d. Yes, the approximation is justified. Our rule of thumb is n  20 and   .05 which is met
here and the probabilities from the Poisson are similar to the binomial.
Learning Objective: 06-6
*6.94
a. P(X = 0). Using Excel: =BINOM.DIST(0,200,.03,0) = .0023
or =POISSON.DIST(0,6,0) = .0025.
b. P(X = 1). Using Excel: =BINOM.DIST(1,200,.03,0) = .0140
or =POISSON.DIST(1,6,0) = .0149.
c. P(X = 2). Using Excel: =BINOM.DIST(2,200,.03,0) = .0430
or =POISSON.DIST(2,6,0) = .0446.
d. Set  = n = (200)(.03) = 6.0 (see parts a through c)
e. Yes, n  20 and   .05 and probabilities are similar.
Learning Objective: 06-6
*6.95
a. For the binomial,  = n = (5708)(.00128) = 7.31 is the expected number of “bumped”
passengers per hour.
b. Using the Poisson approximation:
P(X < 10) = P(X ≤ 9) =POISSON.DIST(9,7.31,1) = .7977.
P(X > 5) =1 − P(X ≤ 5) =1-POISSON.DIST(5,7.31,1) = .7371.
c. Yes, the approximation is justified. Our rule of thumb is n  20 and   .05 which is met.
Learning Objective: 06-6
*6.96
a. For the binomial, µ = n = (500)(.02) = 10 is the expected number of "passes".
b. Using the Poisson approximation with  = 10:
P(X  5) =POISSON.DIST(5,10,1). = .0671.
Learning Objective: 06-6
*6.97
a. Using the Poisson approximation with  = (150)(.025) = 3.75:
P(X ≥ 4) = 1− P(X ≤ 3) =1-POISSON.DIST(3,3.75,1) = .5162
b. Assume calls are independent.
Learning Objective: 06-6
*6.98
a. For the binomial, µ = n = (100,000)(.00014) = 14 is the expected number of "retained
foreign bodies".
b. Using the Poisson approximation with  = n = (100,000)(.00014) = 14 we get
P( X  5) = .0055. The excel function is =POISSON(5,14,1). Less than 1%
chance (.55%) of getting 5 or fewer retained foreign bodies.
Learning Objective: 06-6
115
ASBE 5e Solutions for Instructors
*6.99
a. Geometric mean:  = 1/ = 1/.25 = 4.
b. Using geometric CDF, P(X ≤ 6) = 1 – (1 − π)6 = 1 – (1 − .25)6 = .8220.
Approximately 82.2% change of landing first job offer by the sixth interview.
Learning Objective: 06-9
*6.100
a. Geometric mean is 1/ = 1/(.04) = 25 customers.
b. Using geometric CDF, P( X  20) = 1 − (1 −  ) x = 1 − (1 − .04) 20 = 1–.4420 = .5580.
Approximately 55.8% likelihood of the first birthday cake order coming within
the first 20 customers.
Learning Objective: 06-9
*6.101
a. Geometric mean is 1/ = 1/(.08) = 12.5 cars.
b. Using geometric CDF, P(X  5) = 1−(1−)x = 1− (1−.08)5 = .3409. Approximately 34%
change of finding the first car with a burned out headlight within the first five cars
inspected.
Learning Objective: 06-9
*6.102
a. Geometric mean is 1/ = 1/(.07) = 14.29 operations
b. Using geometric CDF, P(X  20) = 1 − P(X  19) = 1 − [1−(1−)x] = 1–[1–(1−)19]= 1–
(1–.2519) = .2519. Approximately 25% likelihood of conducting 20 or more operations
before the first fatality.
Learning Objective: 06-9
*6.103
a. Geometric mean is 1/ = 1/(.05) = 20.
b. Using geometric CDF, P(X ≥ 30) = 1 − P(X  29) = 1 – (1− (1−)x) = 1 – (1 − (1−.05)29)
= 1 – (1 − .2259) = .2259.
Learning Objective: 06-9
*6.104
Geometric mean is 1/  = 1/(.02) = 50
Learning Objective: 06-9
*6.105
a.  = 233.1 lb 
*6.106
a. By Rule 1, expected total cost is vQ+F = vQ+F = (8)(25000) + 15000 = $350,000
By Rule 2, std dev. of total cost is vQ+F = vQ = (8)(2000) = $16,000
b. To break even, we want TR − TC = 0 where TR = expected total revenue and TC =
expected total cost. Since TR = (Price)(Quantity) = PQ we set PQ − (vQ+F) = 0 and solve
for P to get P(25000) − 350000 = 0 or P = $14. For a profit of $20,000 we have P(25000)
− 370000 = 0 or P = $14.80.
Learning Objective: 06-10
.4536 kg
= 105.73kg
lb
.4536 kg
b.  = 34.95 lb 
= 15.85kg
lb
c. Rule 1 for the mean and Rule 2 for the standard deviation.
Learning Objective: 06-10
116
ASBE 5e Solutions for Instructors
*6.107
 X + X = 1 + 2 = 3420 + 390 = 3810ml ,  X + X =  12 +  2 2 = 102 + 22 = 10.2ml
1
2
1
2
Learning Objective: 06-10
*6.108
a. Using Rule 3: total time = 20+10+14+6+48 = 98 hours
b. Using Rule 4: total time = 42 + 22 + 32 + 22 + 62 = 8.31 . The 2-sigma interval
around the mean is  ± 2 or 98 ± (2)(8.31). The range is 81.4 to 114.6 hours.
Learning Objective: 06-10
*6.109
a. By Rule 1, mean of total cost: vQ+F = vQ+F = (2225)(7) + 500 = $16,075
By Rule 2, std dev. of total cost: vQ+F = vQ = (2225)(2) = $4,450
By Rule 1, expected revenue is E(PQ) = PQ = (2850)(7) = $19,950
Expected profit is TR – TC = 19,950 – 16,075 = $3,875
Learning Objective: 06-10
*6.110
a. total time = 15+30+25+45+20 = 135 minutes, total time = 42 + 62 + 52 + 102 + 52 = 14.2 .
b. Independence might not hold if the outcome of one step affects the duration of the next
procedure.
Learning Objective: 06-10
*6.111
a. X+Y = $70 + $200 = $270
2
2
b. σ X+Y = 10 + 30 + 2  400 = $42.43
c. The variance of the total is greater than either of the individual variances.
Learning Objective: 06-10
117