Chapter 5:
Discrete Random Variables and Probability
Distributions
5.1
Daily computer sales is a discrete random variable that can take on no more than a
countable number of values
5.2
The number of defective parts produced in daily production is a discrete random
variable that can take on no more than a countable number of values.
5.3
a. Discrete – a countable number
b. Discrete - countable
c. Continuous – dollar amounts are generally considered continous, even though
we may truncate dollar amounts and treat dollar amounts as if they were the
same as discrete
d. Discrete - countable
5.4
Discrete random variable – number of plays is countable
5.5
Various answers including; the number of business phone calls on the first day of
business, the number of customers in the first month, the number of employees
hired in the first week, the number of proposals produced for new clients.
5.6
Total sales, advertising expenditures, sales of competitors
5.7
Discrete – the number of voters supporting a candidate is a countable number of
values
5.8
Discrete – the number of purchases is a countable number of values
5.9
Probability distribution of the number of heads
X-number of heads P(x)
0
.5
1
.5
5.10
Probability distribution of number of heads in one toss
X-number of heads P(x)
0
.5
1
.5
Chapter 5: Discrete Random Variables and Probability Distributions
5.11
Probability distribution of number of heads when three coins are tossed
X-number of heads P(x)
0
.125
1
.375
2
.375
3
.125
5.12
Various answers
X –# of times missing class
0
1
2
3
4
5.13
5.14
a.
b.
c.
d.
P(x)
.65
.15
.10
.09
.01
F(x)
.65
.80
.90
.99
1.00
P(3 ≤ x <) = .20 + .20 + .15 = 0.55
P(x > 3) = .20 + .15 + .10 = 0.45
P(x ≤ 4) = .05 + .10 + .20+ .20 + .20 = 0.75
P(2 < x ≤ 5) = .20 + .20 + .15 = 0.55
a. Cumulative probability function:
X
0
1
2
3
4
P(x)
.10 .08 .07 .15 .12
F(x)
.10 .18 .25 .40 .52
5
.08
.60
6
.10
.70
b. P(x ≥ 5) = .08 + .10 + .12 + .08 + .10 = .48
c. P(3 ≤ x ≤ 7) = .15 + .12 + .08 + .10 + .12 = .57
a. Draw the probability distribution function
Chart of Prob_Ex_5.15 vs xi_Ex5.15
0.6
0.5
Prob_Ex_5.15
5.15
0.4
0.3
0.2
0.1
0.0
0
1
xi_Ex5.15
7
.12
.82
8
.08
.90
9
.10
1.00
81
82
Statistics for Business & Economics, 6th Edition
b. Calculate and draw the cumulative probability function
X P(x)
F(x)
0
.40
.40
1
.60
1.00
Chart of Prob_Ex_5.15 vs xi_Ex5.15
Cumulative of Prob_Ex_5.15
1.0
0.8
0.6
0.4
0.2
0.0
0
1
xi_Ex5.15
Cumulative across all data.
c. Find the mean of the random variable of x
X P(x)
0
.40
1
.60
XP(x)
0
.60
.60
x  E( X )   xP( x) = .60
d. Find the variance of X
X P(x)
0
.40
1
.60
 2x
XP(x)
(x-mu)^2
0
.36
.60
.16
.60
 E[( X  x )2 ]   ( x  x )2 P( x)
(x-mu)^2P(x)
.144
.096
.240
= .240
Chapter 5: Discrete Random Variables and Probability Distributions
a. Probability distribution function
Chart of P(X)_Ex5.16 vs xi_Ex5.16
0.5
P(X)_Ex5.16
0.4
0.3
0.2
0.1
0.0
0
1
xi_Ex5.16
2
b. Cumulative probability function
Chart of P(X)_Ex5.16 vs xi_Ex5.16
1.0
Cumulative of P(X)_Ex5.16
5.16
0.8
0.6
0.4
0.2
0.0
0
1
xi_Ex5.16
2
Cumulative across all data.
c. Find the mean
X
0
1
2
P(x)
.25
.50
.25
XP(x)
0
.50
.50
1.00
x  E( X )   xP( x) = 1.00
d. Find the variance of X
X P(x)
XP(x)
(x-mu)^2
0
.25
0
1.0
1
.50
.50
0
2
.25
.50
1
1.00
 2 x  E[( X  x )2 ]   ( x  x )2 P( x) = .50
(x-mu)^2P(x)
.25
0
.25
.50
83
a. Probability distribution function
Chart of P(X)_Ex5.17 vs xi_Ex5.17
0.5
0.4
P(X)_Ex5.17
5.17
Statistics for Business & Economics, 6th Edition
0.3
0.2
0.1
0.0
0
1
xi_Ex5.17
b. Cumulative probability distribution function
Chart of P(X)_Ex5.17 vs xi_Ex5.17
1.0
Cumulative of P(X)_Ex5.17
84
0.8
0.6
0.4
0.2
0.0
0
1
xi_Ex5.17
Cumulative across all data.
c. Find the mean of the random variable of x
X P(x)
XP(x)
0
.50
0
1
.50
.50
.50
x  E( X )   xP( x) = .50
d. Find the variance of X
X P(x)
XP(x)
(x-mu)^2
0
.50
0
.25
1
.50
.50
.25
.50
 2 x  E[( X  x )2 ]   ( x  x )2 P( x)
(x-mu)^2P(x)
.125
.125
.25
= .25
Chapter 5: Discrete Random Variables and Probability Distributions
a. Probability function:
Probability distribution function
Proportion of new cars returned: Correction of defects
Probability: P(x)
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
Number of Returns
b. Cumulative probability function:
Cumulative probability function
Proportion of new cars returned: Correction of defects
Cumulative probability: F(x)
5.18
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0
1
2
3
4
Number of Returns
c.
d.
 = 0 + .36 + 2(.23) + 3(.09) + 4(.04) = 1.25 defects
 2   x2 Px( x)   2 x = 1.1675
85
86
Statistics for Business & Economics, 6th Edition
Excel output:
Returns
0
1
2
3
4
P(x)
0.28
0.36
0.23
0.09
0.04
1.00
F(x)
0.28
0.64
0.87
0.96
1.00
Mean Variance
0
0.4375
0.36
0.0225
0.46 0.129375
0.27 0.275625
0.16
0.3025
1.25
1.1675
S.D.
1.080509
5.19 a. Probability function:
Probability distribution function
New furnace orders resulting from pre-winter service calls
Probability: P(x)
0.3
0.2
0.1
0
1
3
2
Orders
4
5
Chapter 5: Discrete Random Variables and Probability Distributions
b. Cumulative probability function:
Cumulative probability function
New furnace orders resulting from pre-winter service calls
Cumulative probability: F(x)
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
3
2
1
4
5
Orders
c. P(x ≥ 3) = .50
d.  x = 2.45 orders
e.  x = 1.3592 orders
Excel output:
Orders
0
1
2
3
4
5
P(x)
0.10
0.14
0.26
0.28
0.15
0.07
1.00
F(x)
0.10
0.24
0.50
0.78
0.93
1.00
Mean Variance
0 0.60025
0.14 0.29435
0.52 0.05265
0.84
0.0847
0.6 0.360375
0.35 0.455175
2.45
1.8475
S.D.
1.359228
87
88
5.20
Statistics for Business & Economics, 6th Edition
a. Probability function
Probability distribution function
Number of paper clips per package
Probability: P(x)
0.3
0.2
0.1
0.0
47
48
49
50
51
52
53
Clips
b. Cumulative probability function
Cumulative probability: F(x)
Cumulative probability function
Number of paper clips per package
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
47
48
49
50
Clips
c. P(49  x  51) = .70
d. 1 – [P(x < 50)]2 = 1 – .1444 = .8556
51
52
53
Chapter 5: Discrete Random Variables and Probability Distributions
89
e.  x = 47(.04) + 48(.13) + 49(.21) + 50(.29) + 51(.20) + 52(.10) + 53(.03) = 49.9 clips
 2 x = 1.95
 x = 1.3964 clips
Excel output:
f. Mean and standard deviation of profit per package:
 = 1.5 – (.16 + .02X)
 = E() = 1.5 – (.16 + (.02)(49.9)) = $.342
  = |.02|(1.3964) = $.0279
90
Statistics for Business & Economics, 6th Edition
5.21 a. Probability function
Probability distribution function
Number of bus riders
Probability: P(x)
0.3
0.2
0.1
0.0
0
1
2
3
4
5
6
7
Riders
b. Cumulative probability function:
Cumulative probability function
Number of bus riders
Cumulative probability: F(x)
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
Riders
c. P(x  4) = .32
d. P(X < 3 (both days)) = (.37)2 = .1369
5
6
7
Chapter 5: Discrete Random Variables and Probability Distributions
91
e.  = 2.99 riders  2 = 1.9899  = 1.4106 riders
BusRiders
0
1
2
3
4
5
6
7
P(x)
0.02
0.12
0.23
0.31
0.19
0.08
0.03
0.02
1.00
F(x)
0.02
0.14
0.37
0.68
0.87
0.95
0.98
1.00
Mean
0
0.12
0.46
0.93
0.76
0.4
0.18
0.14
2.99
S.D.
Variance
0.17880200
0.47521200
0.22542300
0.00003100
0.19381900
0.32320800
0.27180300
0.32160200
1.98990000
1.41063815
f. Revenue: r = .50X, E(r) = .50(2.99) = 1.495
 r = |.50|(1.4106) = .7053
5.22
a. Probability function
X
0
1
2
P(x)
0.81
0.18
.01
Px(0) = (.90)(.90) = .81
Px(1) = (.90)(.10) + (.10)(.90) = .18
Px(2) = (.10)(.10) = .01
b. P(Y = 0) = 18/20 x 17/19 = 153/190
P(Y=1) = (2/20 x 18/19) + (18/20 x 2/19) = 36/190
P(Y=2) = 2/20 x 1/19 = 1/190
The answer in part b. is different from part a. because in part b. the probability
of picking a defective part on the second draw depends upon the result of the
first draw.
c.  = 0(.81) + .18 + 2(.01) = 0.2 defects
 2 x = .22 – (.20)2 = .18
d.  = 0(153/190) + (36/190) + 2(1/190) = 38/190 = 0.2 defects
 2 y = 40/190 – (.20)2 = .1705
92
Statistics for Business & Economics, 6th Edition
5.23 a. Probability function of X
Px(1) = .40
Px(2) = (.40)(.60) = .24
Px(3) = (.40)(.60)2 = .144
Px(4) = (.40)(.60)3 = .0864
Px(x) = (.40)(.60)x-1 for x = 5, 6, …
b. Cumulative probability function of X
Fx(1) = .40
Fx(2) = .64
Fx(3) = .784
Fx(4) = .8704
Fx(x) = 1 – (.6)x for x = 5, 6, …
c. P(X  3) = 1 – P(X < 3) = 1 - .64 = .36
5.24
“One and one” E(X) = 1(.75)(.25) + 2(.75)2 = 1.3125
“Two-shot foul” E(X) = 1((.75)(.25) + (.25)(.75)) + 2(.75)2 = 1.50
The “two-shot foul” has a higher expected value
5.25
 = E(X) = 0 + .15 + 2(.19) + 3(.26) + 4(.19) + 5(.11) = 2.62 phone calls
 = 1.4470
5.26
 = 3.29
 2 = 1.3259  = 1.1515
Rating
1
2
3
4
5
P(x)
0.07
0.19
0.28
0.30
0.16
1.00
F(x)
0.07
0.26
0.54
0.84
1.00
Mean
0.07
0.38
0.84
1.20
0.80
3.29
S.D.
5.27
 = profit
X = number of requests
S = number of papers stocked
 = .2S - .05(X – S) if X > S
.2X if X = S
.2X - .7(S – X) if X < S
Variance
0.367087
0.316179
0.023548
0.15123
0.467856
1.3259
1.151477
Chapter 5: Discrete Random Variables and Probability Distributions
93
Level of profit for all combinations of S and X:
S\X
0
1
2
3
4
5
0
0.00
-0.70
-1.40
-2.10
-2.80
-3.50
1
-0.05
0.20
-0.50
-1.20
-1.90
-2.60
2
-0.10
0.15
0.40
-0.30
-1.00
-1.70
3
-0.15
0.10
0.35
0.60
-0.10
-0.80
4
-0.20
0.05
0.30
0.55
0.80
0.10
5
-0.25
0.00
0.25
0.50
0.75
1.00
E()|(S=0) = 0 +(-.05)(.16) + (-.1)(.18) + (-.15)(.32) + (-.2)(.14) + (-.25)(.08) = -.122
E()|(S=1) = (-.7)(.12) + (.2)(.16) + (.15)(.18) + (.1)(.32) + (.05)(.14) = .014
E()|(S=2) = (-1.4)(.12) + (-.5)(.16) + (.4)(.18) + (.35)(.32) + (.3)(.14) + (.25)(.08) = -.002
E()|(S=3) = (-2.1)(.12) + (-1.2)(.16) + (-.3)(.18) + (.6)(.32) + (.55)(.14) + (.5)(.08) = .189
E()|(S=4) = (-2.8)(.12) + (-1.9)(.16) + (-1)(.18) + (-.1)(.32) + (.8)(.14) + (.75)(.08) = -.68
E()|(S=5) = (-3.5)(.12) + (-2.6)(.16) + (-1.7)(.18) + (-.8)(.32) + (.1)(.14) + (1)(.08) = 1.304
Store owner maximizes expected profit by ordering one newspaper
5.28
a.  = 1.82 breakdowns
Breakdowns
0
1
2
3
4
P(x)
0.1
0.26
0.42
0.16
0.06
1.00
F(x)
0.10
0.36
0.78
0.94
1.00
Mean
0
0.26
0.84
0.48
0.24
1.82
S.D.
 2 = 1.0276
Variance
0.33124
0.174824
0.013608
0.222784
0.285144
1.0276
1.013706
b. Cost: C = 1500X
E(C) = 1500(1.82) =  = $2,730
 = |1500|(1.0137) = $1,520.559
Cost
0
1500
3000
4500
6000
 = 1.0137 breakdowns
P(x)
0.1
0.26
0.42
0.16
0.06
1.00
F(x)
0.10
0.36
0.78
0.94
1.00
Mean Variance
0 745290
390 393354
1260
30618
720 501264
360 641574
2730 2312100
S.D.
1520.559
94
Statistics for Business & Economics, 6th Edition
5.29
Expected profits are highest for Strategy 1 at $650 vs. $550 for Strategy 2 and
$400 for Strategy 3. The strategy to recommend would depend on the risk
aversion of the investor. The variability of Strategy 1 is much higher than the
variability of Strategy 2. The standard deviation of Strategy 1 is $3,927.7856 vs.
$567.89 for Strategy 2. Many risk averse investors would likely adopt Strategy 2
with its lower standard deviation and hence, lower risk.
5.30
Mean and variance of a Bernoulli random variable with P=.5
x  E( X )   xP( x)  (0)(1  P)  (1) P  P  .5
 2 x  P(1  P)  .5(1  .5)  .25
5.31
Probability of a binomial random variable with P = .5 and n=12, x=7 and x less
than 6
Cumulative Distribution Function
Binomial with n = 12 and p = 0.5
x
0
1
2
3
4
5
6
7
8
9
P( X <= x )
0.000244
0.003174
0.019287
0.072998
0.193848
0.387207
0.612793
0.806152
0.927002
0.980713
Chapter 5: Discrete Random Variables and Probability Distributions
95
P(x=7) = .806152 - .612793 = .1934
P(x<6) = .3872
5.32
Probability of a binomial random variable with P=.3 and n = 14, x=7 and x less
than 6
Cumulative Distribution Function
Binomial with n = 14 and p = 0.3
x
0
1
2
3
4
5
6
7
8
P( X <= x )
0.006782
0.047476
0.160836
0.355167
0.584201
0.780516
0.906718
0.968531
0.991711
P(x=7) = .968531 - .906718 = .06181
P(x<6) = .7805
5.33
Probability of a binomial random variable with P=.4 and n=20, x=9 and x less
than 7
Cumulative Distribution Function
Binomial with n = 20 and p = 0.4
x
0
1
2
3
4
5
6
7
8
9
10
P( X <= x )
0.000037
0.000524
0.003611
0.015961
0.050952
0.125599
0.250011
0.415893
0.595599
0.755337
0.872479
P(x=9) = .755337 - .595599 = .1597
P(x<7) = P(x≤6) = .250011
96
5.34
Statistics for Business & Economics, 6th Edition
Probability of a binomial random variable with P=.7 and n=18, x=12 and x less
than 6
Cumulative Distribution Function
Binomial with n = 18 and p = 0.7
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
P( X <= x )
0.000000
0.000000
0.000000
0.000004
0.000039
0.000269
0.001430
0.006073
0.020968
0.059586
0.140683
0.278304
0.465620
0.667345
P(x=12) = .465620 - .278304 = .1873
P(x<6) = .000269
5.35
Cumulative Distribution Function
Binomial with n = 6 and p = 0.0500000
x
P( X <= x )
0.00
0.7351
1.00
0.9672
2.00
0.9978
3.00
0.9999
4.00
1.0000
5.00
1.0000
a. Px(0) = .7351
b. Px(1) = P(X ≤ 1) – P(X ≤ 0) = .9672 - .7351 = .2321
c. P(X  2) = 1 – P(X ≤ 1) = 1 - .9672 = .0328
5.36
Cumulative Distribution Function
Binomial with n = 5 and p = 0.250000
x
P( X <= x )
0.00
0.2373
1.00
0.6328
2.00
0.8965
3.00
0.9844
4.00
0.9990
5.00
1.0000
a. P(x  1) = 1 – Px(0) = 1 – .2373 = .7627
b. P(x  3) = 1 – P(x ≤ 2) = 1 - .8965 = .1035
Chapter 5: Discrete Random Variables and Probability Distributions
5.37
Cumulative Distribution Function
Binomial with n = 6 and p = 0.700000
x
P( X <= x )
0.00
0.0007
1.00
0.0109
2.00
0.0705
3.00
0.2557
4.00
0.5798
5.00
0.8824
6.00
1.0000
a. P(x  2) = 1 – P(X ≤ 1) = 1 - .0109 = .9891
b. P(x  4) = .5798
5.38
Cumulative Distribution Function
Binomial with n = 7 and p = 0.500000
x
P( X <= x )
0.00
0.0078
1.00
0.0625
2.00
0.2266
3.00
0.5000
4.00
0.7734
5.00
0.9375
6.00
0.9922
7.00
1.0000
P(x  4) = 1 – P(x ≤ 3) = 1 - .5 = .5
5.39
Cumulative Distribution Function
Binomial with n = 6 and p = 0.150000
x
P( X <= x )
0.00
0.3771
1.00
0.7765
2.00
0.9527
3.00
0.9941
4.00
0.9996
5.00
1.0000
6.00
1.0000
a. Px(6) = P(x≤6) – P(x≤5) = 1.0000 – 1.0000  .0000
b. Px(0) = .3771
c. P(X > 1) = 1 – P(X ≤ 1) = 1 - .7765 = .2235
5.40
Cumulative Distribution Function
Binomial with n = 5 and p = 0.400000
x
P( X <= x )
0.00
0.0778
1.00
0.3370
2.00
0.6826
3.00
0.9130
4.00
0.9898
5.00
1.0000
a. P(x = 5) = P(x ≤ 5) – Px ≤ 4) = 1.00 - .9898 = .0102
b. P(x  3) = P(x ≤ 5) – P(x ≤ 2) = 1.000 - .6826 = .3174
97
98
Statistics for Business & Economics, 6th Edition
Cumulative Distribution Function
Binomial with n = 4 and p = 0.400000
x
P( X <= x )
0.00
0.1296
1.00
0.4752
2.00
0.8208
3.00
0.9744
4.00
1.0000
c. P(x  2) = .5248
d. E(X) = np = 5(.4) = 2 games. Unless of course you are a Cubs fan and then
you would hope the Cubs would win all of the games but you would expect
them to win none of the games.
e. E(X) =  = 1 + np = 1 + 4(.4) = 2.6 games
5.41 Find the probability of overbooking a flight. The probability of a ticketed passenger
showing up for a flight is 1 - .2 = .8. Therefore, based on n = 10 tickets sold and a
probability (p) of the ticketed passenger showing up, the probabilities of the
binomial distribution are shown below.
Probability Density Function
Binomial with n = 10 and p = 0.8
x P( X = x )
0
0.000000
1
0.000004
2
0.000074
3
0.000786
4
0.005505
5
0.026424
6
0.088080
7
0.201327
8
0.301990
9
0.268435
10
0.107374
Since 10% of the time 9 tickets are sold and 5% of the time 10 tickets are sold, the
proportion of flights where the number of ticketed passengers showing up exceeds
the number of available seats is: (.10)(.268435) + (.05)(.107374) = .0322.
5.42
Cumulative Distribution Function
Binomial with n = 4 and p = 0.400000
x
P( X <= x )
0.00
0.1296
1.00
0.4752
2.00
0.8208
3.00
0.9744
4.00
1.0000
a. P(x  2) = P(X ≤ 4) – P(X ≤ 1) = 1.000 - .4752 = .5248
b. E(X) = np = 4(.4) = 1.6  x  4(.4)(.6)  .9798
Chapter 5: Discrete Random Variables and Probability Distributions
5.43
a E(X) = 50(.15) = 7.5
 x  50(.15)(.85)  2.5249
b. Let Z = 250X
E(Z) = 250(7.5) = $1,875
 z = |250|(2.5249) = $631
5.44
a. E(X) = 2000(.032) = 64
 x  2000(.032)(.968)  7.871
b. Let Z = 10X
E(Z) = 10(64) = $640
 z = |10|(7.871) = $78.71
5.45
 x = 2.0 sales
OR
 x = np = 5(.4) = 2.0 sales
5.46
a. E(X) =  x = np = 620(.78) = 483.6,  x =
b. Let Z = 2X
E(Z) = 2(483.6) = $967.20
 z = |2|(10.314) = $20.6292
5.47
a. Px(0) + Px(1) = (.95)16 + 16(.05)(.95)15 = .8108
b. Px(0) + Px(1) = (.85)16 + 16(.15)(.85)15 = .2839
c. Px(0) + Px(1) = (.75)16 + 16(25)(.75)15 = .0635
620(.78)(.22) = 10.3146
99
100
5.48
Statistics for Business & Economics, 6th Edition
The acceptance rules have the following probabilities:
(i) Rule 1: P(X=0) = (.8)10 = .1074
(ii) Rule 2: P(X ≤ 1) = (.8)20 + 20(.2)(.8)19 = .0692
Therefore, the acceptance rule with the smaller probability of accepting a
shipment containing 20% defectives will be the second acceptance rule
(( 201 )(.1)(.9)19 )(.7)
P(Supplier1|x=1) = 20
 .916
(( 1 )(.1)(.9)19 )(.7)  (( 201 )(.2)(.8)19 )(.3)
5.49
5.50
Probability Density Function
Hypergeometric with N = 50, M = 25, and n = 12
x
0
1
2
3
4
5
P( X = x )
0.000043
0.000918
0.008078
0.038706
0.112702
0.210376
P(x=5) = .210376
5.51
Probability Density Function
Hypergeometric with N = 60, M = 25, and n = 14
x
0
1
2
3
4
5
6
7
P( X = x )
0.000134
0.002128
0.014432
0.055323
0.133881
0.216269
0.240299
0.186354
P(x=7) = .186354
5.52
Probability Density Function
Hypergeometric with N = 80, M = 42, and n = 20
x
0
1
2
3
4
5
6
7
8
9
P( X = x )
0.000000
0.000000
0.000008
0.000093
0.000704
0.003723
0.014348
0.041322
0.090392
0.151769
P(x=9) = .151769
Chapter 5: Discrete Random Variables and Probability Distributions
101
5.53
Probability Density Function
Hypergeometric with N = 40, M = 25, and n = 5
x
0
1
2
3
P( X = x )
0.004564
0.051861
0.207444
0.367017
P(x=3) = .367017
5.54
Probability Density Function
Hypergeometric with N = 400, M = 200, and n = 15
x
0
1
2
3
4
5
6
7
8
P( X = x )
0.000023
0.000375
0.002792
0.012743
0.039848
0.090434
0.153879
0.199906
0.199906
P(x=8) = .1999
5.55
a. P(Shipment is accepted) can be found by: P(x = 0) with N=16, S=4, n = 4: =
.2720.
Cumulative Distribution Function
Hypergeometric with N = 16, X = 4, and n = 4
x
P( X <= x )
0.00
0.2720
1.00
0.7555
2.00
0.9731
3.00
0.9995
4.00
1.0000
b. P(Shipment is accepted) can be found by: P(x = 0) with N=16, S=1, n = 4: =
.7500
Cumulative Distribution Function
Hypergeometric with N = 16, X = 2, and n = 4
x
P( X <= x )
0.00
0.5500
1.00
0.9500
2.00
1.0000
3.00
1.0000
4.00
1.0000
102
Statistics for Business & Economics, 6th Edition
c. P(Shipment is rejected) can be found by taking 1 minus the P(Shipment is
accepted): [1-P(x = 0)] with N=16, S=1, n = 4: = [1 - .75] = .25
Cumulative Distribution Function
Hypergeometric with N = 16, X = 1, and n = 4
x
P( X <= x )
0.00
0.7500
1.00
1.0000
2.00
1.0000
3.00
1.0000
4.00
1.0000
5.56
Cumulative Distribution Function
Hypergeometric with N = 16, X = 8, and n = 8
x
P( X <= x )
1.00
0.0051
2.00
0.0660
3.00
0.3096
4.00
0.6904
5.00
0.9340
6.00
0.9949
7.00
0.9999
P(x = 4) = P(x  4) – P(x  3) = .6904 - .3096 = .3808
5.57
Cumulative Distribution Function
Hypergeometric with N = 12, X = 4, and n = 3
x
P( X <= x )
0.00
0.2545
1.00
0.7636
2.00
0.9818
3.00
1.0000
P(x  2) = 1 – P(x  1) = 1 - .7636 = .2364
5.58
Cumulative Distribution Function
Hypergeometric with N = 10, X = 5, and n = 6
x
P( X <= x )
0.00
0.0000
1.00
0.0238
2.00
0.2619
3.00
0.7381
4.00
0.9762
5.00
1.0000
P(x  2) = .2619
Chapter 5: Discrete Random Variables and Probability Distributions
5.59
Probability Density Function
Poisson with mean = 3.5
x
0
1
2
3
4
5
6
7
P( X = x )
0.030197
0.105691
0.184959
0.215785
0.188812
0.132169
0.077098
0.038549
P(x=7) = .038549
5.60
Probability Density Function
Poisson with mean = 2.5
x
0
1
2
3
4
P( X = x )
0.082085
0.205212
0.256516
0.213763
0.133602
P(x=4) = .1336
5.61
Cumulative Distribution Function
Poisson with mean = 4.5
x
0
1
2
3
4
5
6
7
8
P( X <= x )
0.011109
0.061099
0.173578
0.342296
0.532104
0.702930
0.831051
0.913414
0.959743
P(x>7) = 1-(Px≤ 7) = 1 - .913414 = .086586
5.62
Cumulative Distribution Function
Poisson with mean = 3.5
x
0
1
2
3
4
5
6
P( X <= x )
0.030197
0.135888
0.320847
0.536633
0.725445
0.857614
0.934712
P(x<6) = .857614
103
104
Statistics for Business & Economics, 6th Edition
5.63
Cumulative Distribution Function
Poisson with mean = 8
x
0
1
2
3
4
5
6
7
8
9
P( X <= x )
0.000335
0.003019
0.013754
0.042380
0.099632
0.191236
0.313374
0.452961
0.592547
0.716624
P(x≤9) = .716624
5.64
Cumulative Distribution Function
Poisson with mu = 3.00000
x
P( X <= x )
0.00
0.0498
1.00
0.1991
2.00
0.4232
3.00
0.6472
4.00
0.8153
5.00
0.9161
6.00
0.9665
7.00
0.9881
8.00
0.9962
9.00
0.9989
10.00
0.9997
P(x  2) = .4232
5.65
a. P(x < 2) = P(x  1) = .2674
b. P(x > 3) = 1 – P(x  3) = 1 - .7360 = .2640
Cumulative Distribution Function
Poisson with mu = 2.60000
x
P( X <= x )
0.00
0.0743
1.00
0.2674
2.00
0.5184
3.00
0.7360
4.00
0.8774
5.00
0.9510
6.00
0.9828
7.00
0.9947
8.00
0.9985
9.00
0.9996
10.00
0.9999
Chapter 5: Discrete Random Variables and Probability Distributions
5.66
Cumulative Distribution Function
Poisson with mu = 4.20000
x
P( X <= x )
0.00
0.0150
1.00
0.0780
2.00
0.2102
3.00
0.3954
4.00
0.5898
5.00
0.7531
6.00
0.8675
7.00
0.9361
8.00
0.9721
9.00
0.9889
10.00
0.9959
P(x  3) = 1 – P(x  2) = 1 - .2102 = .7898
5.67
Cumulative Distribution Function
Poisson with mu = 3.20000
x
P( X <= x )
0.00
0.0408
1.00
0.1712
2.00
0.3799
3.00
0.6025
4.00
0.7806
5.00
0.8946
6.00
0.9554
7.00
0.9832
8.00
0.9943
9.00
0.9982
10.00
0.9995
a. P(x < 2) = P(x  1) = .1712
b. P(x > 4) = 1 – P(x  4) = 1 - .7806 = .2194
5.68
Cumulative Distribution Function
Poisson with mu = 5.50000
x
P( X <= x )
0.00
0.0041
1.00
0.0266
2.00
0.0884
3.00
0.2017
4.00
0.3575
5.00
0.5289
6.00
0.6860
7.00
0.8095
8.00
0.8944
9.00
0.9462
10.00
0.9747
P(x  2) = .0884
105
106
Statistics for Business & Economics, 6th Edition
5.69
Cumulative Distribution Function
Poisson with mu = 2.50000
x
P( X <= x )
0.00
0.0821
1.00
0.2873
2.00
0.5438
3.00
0.7576
4.00
0.8912
5.00
0.9580
6.00
0.9858
7.00
0.9958
8.00
0.9989
9.00
0.9997
10.00
0.9999
P(x < 4) = P(x  3) = .7576
5.70
Cumulative Distribution Function
Poisson with mu = 6.00000
x
P( X <= x )
0.00
0.0025
1.00
0.0174
2.00
0.0620
3.00
0.1512
4.00
0.2851
5.00
0.4457
6.00
0.6063
7.00
0.7440
8.00
0.8472
9.00
0.9161
10.00
0.9574
P(x  3) = 1 - P(x  2) = 1 - .0620 = .9380
5.71
Cumulative Distribution Function
Poisson with mu = 4.50000
x
P( X <= x )
0.00
0.0111
1.00
0.0611
2.00
0.1736
3.00
0.3423
4.00
0.5321
5.00
0.7029
6.00
0.8311
7.00
0.9134
8.00
0.9597
9.00
0.9829
10.00
0.9933
P( x  3) = 1 - P(x  2) = 1 - .1736 = .8264
The calculations to find the exact binomial probabilities would be to use the
binomial formula for each of the individual probabilities: P(3) + P(4) + P(5) +
P(6) + …+ P(60). Thus, the binomial formula would need to be utilized 58 times
to calculate the exact probability.
Chapter 5: Discrete Random Variables and Probability Distributions
107
Two models are possible – the poisson distribution is appropriate when the
warehouse is serviced by many thousands of independent truckers where the
mean number of ‘successes’ is relatively small. However, under the assumption
of a small fleet of 10 trucks with a probability of any truck arriving during a
given hour is .1, then the binomial distribution is the more appropriate model.
Both models yield similar, although not identical, probabilities.
5.72
Cumulative Distribution Function
Poisson with mean = 1
x
0
1
2
3
4
5
6
7
8
9
10
P( X <= x )
0.36788
0.73576
0.91970
0.98101
0.99634
0.99941
0.99992
0.99999
1.00000
1.00000
1.00000
Cumulative Distribution Function
Binomial with n = 10 and p = 0.1
x
0
1
2
3
4
5
6
7
8
9
10
5.73
P( X <= x )
0.34868
0.73610
0.92981
0.98720
0.99837
0.99985
0.99999
1.00000
1.00000
1.00000
1.00000
a. Compute marginal probability distributions for X and Y
Exercise_5.73
Y_5.73
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
1
0.25
0.25
0.5
X_5.73
2
0.25
0.25
0.5
0.5
0.125
1
0.125
1.5
0.25
0.5
0.25
0.5
0.75
0
P(y) Mean of Y
0.5
0
0.5
0.5
1
0.5
Var of Y StDev of Y
0.125
0.125
0.25
0.5
108
Statistics for Business & Economics, 6th Edition
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .75 – (1.5)(.5) = 0.0
x
y
  Corr ( X , Y ) 
Cov( X , Y )
 x y
= 0.0/(.5)(.5) = 0.0
Note that when covariance between X and Y is equal to zero, it follows that the
correlation between X and Y is also zero.
5.74
a. Compute marginal probability distributions for X and Y
Exercise_5.74
Y_5.74
1
0.2
0.3
0.5
X_5.74
2
0.25
0.25
0.5
0.5
0.125
1
0.125
1.5
0.25
0.5
0.3
0.5
0.8
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
P(y) Mean of Y Var of Y StDev of Y
0.45
0 0.136125
0.55
0.55 0.111375
1
0.55
0.2475 0.497494
-0.025
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .80 – (1.5)(.55) = -.025
x
y
  Corr ( X , Y ) 
5.75
Cov( X , Y )
 x y
= -.025/(.5)(.497494) = -.1005
a. Compute marginal probability distributions for X and Y
Exercise_5.75
Y_5.75
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
1
0.25
0.25
0.5
X_5.75
2
0.25
0.25
0.5
0.5
0.125
1
0.125
1.5
0.25
0.5
0.25
0.5
0.75
0
P(y) Mean of Y
0.5
0
0.5
0.5
1
0.5
Var of Y StDev of Y
0.125
0.125
0.25
0.5
Chapter 5: Discrete Random Variables and Probability Distributions
109
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .75 – (1.5)(.5) = 0.0
x
y
  Corr ( X , Y ) 
Cov( X , Y )
 x y
= 0.0/(.5)(.5) = 0.0
Note that when covariance between X and Y is equal to zero, it follows that the
correlation between X and Y is also zero.
c. Compute the mean and variance for the linear function W = X + Y
W  a  x  b y = (1)1.5 + (1).5 = 2.0
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  12 (.25)  12 (.25)  2(1)(1)(0.0)  .50
5.76
a. Compute marginal probability distributions for X and Y
Exercise_5.76
Y_5.76
1
0.3
0.25
0.55
X_5.76
2 P(y)
0.2
0.25
0.45
Mean of X
Var of X
StDev of X
0.55
0.55
0.9
1.8
1.45
2.35
1.532971
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
0.25
0.5
0.75
0
1
P(x)
Mean of Y Var of Y StDev of Y
0.5
0
0.125
0.5
0.5
0.125
1
0.5
0.25
0.5
0.025
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .75 – (1.45)(.5) = 0.025
x
  Corr ( X , Y ) 
y
Cov( X , Y )
 x y
= 0.025/(1.53297)(.5) = 0.0326
c. Compute the mean and variance for the linear function W = 2X + Y
W  a  x  b y = (2)1.45 + (1).5 = 3.4
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  22 (2.35)  12 (.25)  2(2)(1)(0.025)  9.75
110
5.77
Statistics for Business & Economics, 6th Edition
a. Compute marginal probability distributions for X and Y
Exercise_5.77
Y_5.77
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
1
0.7
0
0.7
X_5.77
2
0
0.3
0.3
0.7
0.063
0.6
0.147
1.3
0.21
0.458258
0
0.6
0.6
P(y) Mean of Y
0.7
0
0.3
0.3
1
0.3
Var of Y StDev of Y
0.063
0.147
0.21 0.458258
0.21
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .60 – (1.3)(.3) = 0.21
x
y
  Corr ( X , Y ) 
Cov( X , Y )
 x y
= 0.21/(.458258)(.458258) = 1.00
c. Compute the mean and variance for the linear function W = 3X + 4Y
W  a  x  b y = (3)1.3 + (4).3 = 5.1
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  32 (.21)  42 (.21)  2(3)(4)(1.0)  29.25
5.78
a. Compute the marginal probability distributions for X and Y
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
1
0.25
0.25
0.5
2 P(y)
0.25 0.5
0.25 0.5
0.5
1
0.5
1 1.5
0.125 0.125 0.25
0.5
0.25
0
0.5 0.75
Mean of Y
0
0.5
0.5
Var of Y
0.125
0.125
0.25
StDev of Y
0.5
Chapter 5: Discrete Random Variables and Probability Distributions
111
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .75 – (1.5)(.5) = 0.0
x
y
  Corr ( X , Y ) 
Cov( X , Y )
 x y
= 0.0/(.5)(.5) = 0.0
Note that when covariance between X and Y is equal to zero, it follows that the
correlation between X and Y is also zero.
c. Compute the mean and variance for the linear function W = X - Y
W  a  x  b y = (1)1.5 + (-1).5 = 1.0
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  12 (.25)  12 (.25)  2(1)( 1)(0.0)  .5
5.79
a. Compute the marginal probability distributions for X and Y.
1
2
0.3 0.2
0.25 0.25
0.55 0.45
0
1
P(x)
P(y)
0.5
0.5
1
Mean of X
Var of X
StDev of X
0.55
0.55
0.9
1.8
1.45
2.35
1.532971
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
0.25
0.5
0.75
Mean of Y Var of Y
0
0.125
0.5
0.125
0.5
0.25
StDev of Y
0.5
0.025
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .75 – (1.45)(.5) = 0.025
x
  Corr ( X , Y ) 
y
Cov( X , Y )
 x y
= 0.025/(1.53297)(.5) = 0.0326
c. Compute the mean and variance for the linear function W = 2X - Y
W  a  x  b y = (2)1.45 + (-1).5 = 2.4
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  22 (2.35)  12 (.25)  2(2)(1)(0.025)  9.55
112
5.80
Statistics for Business & Economics, 6th Edition
a. Compute the marginal probability distributions for X and Y.
Exercise_5.80
Y_5.80
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
1
0
0.4
0.4
X_5.80
2
0.6
0
0.6
0.4
0.144
1.2
0.096
1.6
0.24
0.489898
0.4
0
0.4
P(y) Mean of Y Var of Y StDev of Y
0.6
0
0.096
0.4
0.4
0.144
1
0.4
0.24 0.489898
-0.24
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .40 – (1.6)(.4) = -0.24
x
y
  Corr ( X , Y ) 
Cov( X , Y )
 x y
= -0.24/(.489898)(.489898) = -1.00
c. Compute the mean and variance for the linear function W = 2X - 4Y
W  a  x  b y = (2)1.6 + (-4).4 = 1.6
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  22 (.24)  (4) 2 (.24)  2(2)(4)(.24)  8.64
5.81
a. Compute marginal probability distributions for X and Y
0
1
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
Cov(x,y) =
sum xyP(x)-muxmuy
1
0.7
0
0.7
2
0
0.3
0.3
P(y)
0.7
0.3
1
0.7
0.6
0.063 0.147
1.3
0.21
0.458258
0
0.21
0.6
0.6
Mean of Y Var of Y
0
0.063
0.3
0.147
0.3
0.21
StDev of Y
0.458258
Chapter 5: Discrete Random Variables and Probability Distributions
113
b. Compute the covariance and correlation for X and Y
Cov( X , Y )    xyP( x, y )   x  y = .60 – (1.3)(.3) = 0.21
x
  Corr ( X , Y ) 
y
Cov( X , Y )
 x y
= 0.21/(.458258)(.458258) = 1.00
c. Compute the mean and variance for the linear function W = 10X - 8Y
W  a  x  b y = (10)1.3 + (-8).3 = 10.6
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  102 (.21)  (8) 2 (.21)  2(10)(8)(.21)  .84
5.82 a. Px(0) = .07 + .07 + .06 + .02 = .22
Px(1) = .09 + .06 + .07 + .04 = .26
Px(2) = .06 + .07 + .14 + .16 = .43
Px(3) = .01 + .01 + .03 + .04 = .09
 x = 0 + .26 + 2(.43) + 3(.09) = 1.39
b. Py(0) = .07 + .09 + .06 + .01 = .23
Py(1) = .07 + .06 + .07 + .01 = .21
Py(2) = .06 + .07 + .14 + .03 = .30
Py(3) = .02 + .04 + .16 + .40 = .26
 y = 0 + .21 + 2(.3) + 3(.26) = 1.59
c. PY|X(0|3) = .01/.09 = .1111
PY|X(1|3) = .01/.09 = .1111
PY|X(2|3) = .03/.09 = .3333
PY|X(3|3) = .04/.09 = .4444
d. Cov( X , Y )  E ( XY )   x  y
E(XY) = 0 + 1(1)(.06) + 1(2)(.07) + 1(3)(.04) + 2(1)(.07) + 2(2)(.14)
+ 2(3)(.16) + 3(1)(.01) + 3(2)(.03) + 3(3)(.04) = 2.55
Cov( X , Y )  2.55  (1.39)(1.59) = .3399
e. No, because Cov( X , Y )  0
5.83
a. Joint cumulative probability function at X = 1, Y = 4:
FX,Y(1,4) = .09 + .07 + .14 + .23 = .53
b. PY|X(3|0) = .09/.19 = .4737
PY|X(4|0) = .07/.19 = .3684
PY|X(5|0) = .03/.19 = .1579
114
Statistics for Business & Economics, 6th Edition
c. PY|X(0|5) = .03/.24 = .125
PY|X(1|5) = .10/.24 = .4167
PY|X(2|5) = .11/.24 = .4583
d. E(XY) = 0 + 1(3)(.14) + 1(4)(.23) + 1(5)(.10) + 2(3)(.07) + 2(4)(.16) +
2(5)(.11) = 4.64
 x  0  .47  2(.34)  1.15
 y  3(.3)  4(.46)  5(.24)  3.94
Cov( X , Y )  4.64  (1.15)(3.94) = .109
The covariance indicates that there is a positive association between the
number of lines in the advertisement and the volume of inquiries.
e. No, because Cov( X , Y )  0
X Return
3
4
5
P(y)
Mean of Y
Var of Y
StDev of Y
xyP(x)
1
0.14
0.23
0.1
0.47
0
0.47
0.68
0.251275 0.010575 0.24565
0
sum xyP(x)*muxmuy
5.84
0
0.09
0.07
0.03
0.19
Y Return
2
0.07
0.16
0.11
0.34
1.84
2.8
P(x) Mean of X Var of X StDev of X
0.3
0.9 0.26508
0.46
1.84 0.001656
0.24
1.2 0.269664
3.94
0.5364 0.732393
1.15
0.5075
0.4956309
4.64
0.109
a. Py(0) = .08 + .03 + .01 = .12
Py(1) = .13 + .08 + .03 = .24
Py(2) = .09 + .08 + .06 = .23
Py(3) = .06 + .09 + .08 = .23
Py(4) = .03 + .07 + .08 = .18
b. PY|X(y|3) = 1/26; 3/26; 6/26; 8/26; 8/26
c. No, because Px,y(3,4) = .08 ≠ .0468 = Px(3)Py(4)
Chapter 5: Discrete Random Variables and Probability Distributions
5.85
a. P(0,0)=.54, P(0,1)=.30, P(1,0)=.01, P(1,1)=.15
b. PY|X(y|1) = 1/16 = .0625; 15/16 = .9375
c. E(XY) = .15
 x  0  1(.16)  .16
 y  0  1(.45)  .45
Cov( X , Y )  .15  (.16)(.45)  .078
The covariance indicates that there is a positive association between brand
watchers of a late-night talk show and brand name recognition.
X Watch
0
0.54
0.01
0.55
0
1
P(y)
Mean of Y
Var of Y
StDev of Y
0
Sum xyP(x)*muxmuy
a.
Y/X
0
1
Total
1
0.3
0.15
0.45
0
0.45
0.111375 0.136125
xyP(x)
5.86
115
0
.704
.096
.80
0.15
Y Identify
P(x) Mean of X Var of X
StDev of X
0.84
0 0.021504
0.16
0.16 0.112896
0.16
0.1344 0.366606056
0.45
0.2475
0.49749372
0.15
0.078
1
.168
.032
.20
Total
.872
.128
1.00
b. PY|X(y|0) = .88; .12
c. Px(0) = .80
Px(1) = .20
Py(0) = .872
Py(1) = .128
d. E(XY) = .032;
 x  0  1(.20)  .20 ,  y  0  1(.128)  .128
Cov( X , Y )  .032  (.20)(.128)  .0064
The covariance indicates that there is a positive association between X and Y,
professors are more likely to be away from the office on Friday than during the
other days.
116
Statistics for Business & Economics, 6th Edition
5.87
Because of independence, the joint probabilities are the products of the marginal
probabilities, so P(0,0)=.0216, and so on.
X Food
0
1
2
3
P(y)
Mean of Y
Var of Y
StDev of Y
0
0.0216
0.0522
0.0756
0.0306
0.18
1
0.0456
0.1102
0.1596
0.0646
0.38
Y Service
2
0.0408
0.0986
0.1428
0.0578
0.34
3
0.012
0.029
0.042
0.017
0.1
0
0.38
0.68
0.3
0.332928 0.049248 0.139264 0.26896
P(x) Mean of X Var of X StDev of X
0.12
0 0.322752
0.29
0.29 0.118784
0.42
0.84 0.054432
0.17
0.51 0.314432
1
1.64
0.8104 0.900222
1.36
0.7904
0.889044
5.88 See table above. Number of total complaints (food complaints + service
complaints) has a mean of (1.36 + 1.64) = 3.00. If the two types of complaints are
independent, then the variance of total complaints is equal to the sum of the variance
of the two types of complaints because the covariance would be zero. (.8104 +
.7904) = 1.6008. The standard deviation will be the square root of the variance =
1.26523.
If the number of food and service complaints are not independent of each other, then
the covariance would no longer be zero. The mean would remain the same; however,
the standard deviation would change. The variance of the sum of the two types of
complaints becomes the variance of one plus the variance of the other plus two times
the covariance.
5.89
Y Small
X Large
0
1
2
3
4
5
P(y)
Mean of Y
Var of Y
StDev of Y
0
0.0144
0.0288
0.0504
0.0576
0.018
0.0108
0.18
1
0.0208
0.0416
0.0728
0.0832
0.026
0.0156
0.26
0
0.26
0.49005 0.10985
2
0.0288
0.0576
0.1008
0.1152
0.036
0.0216
0.36
3
0.0104
0.0208
0.0364
0.0416
0.013
0.0078
0.13
4 P(x)
0.0056
0.0112
0.0196
0.0224
0.007
0.0042
0.07
0.72
0.39
0.28
0.0441 0.23693 0.38658
Mean
of X
0.08
0.16
0.28
0.32
0.1
0.06
1
1.65
1.2675
1.12583302
StDev of
Var of X X
0 0.453152
0.16 0.304704
0.56 0.040432
0.96 0.123008
0.4 0.26244
0.3 0.411864
2.38
1.5956 1.263171
Chapter 5: Discrete Random Variables and Probability Distributions
117
  5 x  10 y = 5(2.38) + 10(1.65) = 28.4
  (5) x 2  (10) y 2  5(1.5965)  10(1.2675) = 4.545
5.90
a. No, not necessarily. There is a probability distribution associated with the rates
of return in the mutual fund and not all rates of return will equal the expected
value.
b.Which fund to invest in will depend not only on the expected value of the return
but also on the riskiness of each fund and how risk averse the client is.
5.91
Days
1
2
3
4
5
Ex 5.56
P(x)
0.05
0.2
0.35
0.3
0.1
1.00
F(x)
0.05
0.25
0.60
0.90
1.00
Mean Variance
0.05
0.242
0.4
0.288
1.05
0.014
1.2
0.192
0.5
0.324
3.2
1.06
S.D.
1.029563
a.
b.
c.
d.
P(x < 3) = .05+ .20 = .25
E(X) = 3.2
 = 1.029563
Cost = $20,000 + $2,000X = E(Cost) = $26,400,
standard deviation = ($2,000)(1.029563) = $2,059.13
e. The probability of a project taking at least 4 days to complete is .30 + .10 =
.4. Given independence of the individual projects, the probability that at
least two of three projects will take at least 4 days to complete is a binomial
random variable with n = 3, p = .4. P(2) + P(3) = 3(.4)2(.6) + (1)(.4)3(1) =
.352
5.92
Cars
0
1
2
3
4
5
Ex 5.57
P(x)
0.1
0.2
0.35
0.16
0.12
0.07
1.00
F(x)
0.10
0.30
0.65
0.81
0.93
1.00
Mean
0
0.2
0.7
0.48
0.48
0.35
2.21
S.D.
Variance
0.48841
0.29282
0.015435
0.099856
0.384492
0.544887
1.8259
1.351259
a. E(X) = 2.21 cars sold
b. Standard deviation = 1.3513 cars
c. Mean Salary = $250 + $300 (2.21) = $913. Standard deviation of salary =
$300(1.3513) = $405.39
118
Statistics for Business & Economics, 6th Edition
d. To earn a salary of $1,000 or more, the salesperson must sell at least 3 cars.
P(X  3) = .16 + .12 + .07 = .35
5.93
a.  = np = 9(.25) = 2.25
b)   np(1  p) = 9(.25)(.75) = 1.299
c) (i) E(X) = 1 + 2.23 = 3.25, (ii)  = 1.299
5.94
a. Positive covariance: Consumption expenditures & Disposable income
b. Negative covariance: Price of cars and the number of cars sold
c. Zero covariance: Dow Jones stock market average & rainfall in Brazil
5.95
a. P(4) = (.95)(.90)(.90)(.80) = .6156
P(3) = (.05)(.90)(.90)(.8) + 2(.95)(.10)(.90)(.80) + (.95)(.90)(90)(.20) =
.3231
P(2) = 2(.95)(.90)(.10)(.2) + 2(.05)(.90)(.10)(.80) + (.05)(.90)(.90)(.2) +
(.95)(.10)(.10)(.8) = .0571
P(1) = (.95)(.10)(.10)(.2) + 2(.05)(.90)(.10)(.20) + (.05)(.10)(.10)(.8) =
.0041
P(0) = (.05)(.10)(.10)(.20) = .0001
b. E(X) = .0041 + 2(.0571) + 3(.3231) + 4(.6156) = 3.55 vehicles
c.
 x Px( x)  12.99 , 
2
x
 12.99  (3.55) 2  .6225 vehicles
5.96
X Years
Y Visits
0
1
2
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
sum xyP(x)*muxmuy
1
0.07
0.13
0.04
0.24
4
0.02
0.15
0.1
0.27
P(y)
0.17
0.56
0.27
1
0.24
0.4
0.87
1.08
0.606744 0.0696 0.048749 0.5368
2.59
1.2619
1.12334
0.21
0.191
2
0.05
0.11
0.04
0.2
0.38
3
0.03
0.17
0.09
0.29
1.05
1.4
3.04
Mean Var of StDev of
of Y
Y
Y
0 0.2057
0.56 0.0056
0.54 0.2187
1.1
0.43 0.6557439
Chapter 5: Discrete Random Variables and Probability Distributions
119
a. Py(0) = .07+.05+.03+.02 = .17
b. E(X) =  x  .24  2(.2)  3(.29)  4(.27)  2.59
E(Y) =  y  .56  2(.27)  1.1
c. E(XY)=3.04, Cov(X,Y) = 3.04 – (2.59)(1.1) = .191. This implies that there
is a positive relationship between the number of years in school and the
number of visits to a museum in the last year.
5.97
Assume that the shots are independent of each other
6
6
a.
P(x  2) = 1 – P(x  1) = 1 – [(   (.4)0 (.6)6    (.4)(.6)5 ] = 0.767
0
1 
b.
c.
d.
5.98
6
P(x=3) =   (.4)3 (.6)3 = 0.2765
3
  np = (6)(.4) = 2.4,   6(.4)(.6) = 1.2
Mean of total points scored = 3() = 3(2.4) = 7.2, Std dev = 3 =
3(1.2)=3.6
 5
a. P(x=3) =   .553.452 = .3369
 3
b. P(x  3) = P(3)+P(4)+P(5) = .3369 + (5)(.55)4(.45) + (1)(.55)5(1) = .5931
c.   np = (80)(.55) = 44 will graduate in 4 years. The proportion is 44/80 =
.55.   80(.55)(.45) = 4.4497. The proportion is 4.4497/80 = .05562
5.99
a. This is a binomial probability (assuming independence) with a p=.6 and n=7.
7
7
Then the P(A wins) = P(X ≥ 4) =   .64.43    .65.42  7(.66 ).4  .67 =
 4
5
0.71021
6
 
b.   .63.43 = 0.27648
3
c. (i) The outcome of the first four games are known with certainty.
Therefore, the series is a best out of three games. To compute the
probability that team A wins, find P(x  3) = 3(.6)2(.4) + (.6)3 = 0.648,
 2
(ii)   (.6)(.4) = 0.48
1 
120
5.100
5.101
Statistics for Business & Economics, 6th Edition
To evaluate the effectiveness of the analyst’s ability, find the probability that x
is greater than or equal to 3 at random. P(x  3) =
 5  10   5  10   5  10 
        
 3  2    4  1    5  0  = .16683
15 
 15 
 15 
 
 
 
5 
5 
5 
C04C416  C14C316
1820  2240
Find P(X  2) = 1 = .16202
 1
20
C4
4845
5.102
a. P(0) = e-2.4 = .09072
b. P(x > 3) = 1 – e-2.4 - e-2.4(2.4) - e-2.4 (2.4)2/2! - e-2.4 (2.4)3/3! = .2213
5.103
1 – e-6.5 - e-6.5(6.5) - e-6.5 (6.5)2/2! = 0.95696
5.104 P(x=0) = e-2.4 = .0907
Let Y be the number of stalls for both lines.
Find the P(Y  1) = 1 – P(Y=0) = 1 – (.0907)2 = .99177
5.105 compute the mean and variance for the total value of the stock portfolio
Exercise_5.105
Y_5.105
45
50
55
60
P(x)
Mean of X
Var of X
StDev of X
xyP(x)
cov(x,y) =
sum xyP(x)-muxmuy
40
0
0.05
0.1
0.2
0.35
X_5.105
50
0
0
0.05
0.1
0.15
60
0.05
0.05
0
0.05
0.15
14
78.75
7.5
3.75
9
3.75
24.5
78.75
55
165
12.84523
900
437.5
615
1522.5
3475
70 P(y)
Mean of Y Var of Y StDev of Y
0.2
0.25
11.25 17.01563
0.1
0.2
10
2.1125
0.05
0.2
11
0.6125
0
0.35
21 15.94688
0.35
1
53.25 35.6875 5.973902
546.25
Compute the mean and variance for the linear function W = X + Y
W  a  x  b y = (1)55 + (1)53.25 = 108.25
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  12 (165)  12 (35.6875)  2(1)(1)(546.25)  1293.1875
Chapter 5: Discrete Random Variables and Probability Distributions
121
5.106 Compute the mean and variance
Exercise_5.106
Y_5.106
3
0.1
0.1
0.05
X_5.106
4
0.15
0.2
0.15
P(x)
0.25
0.5
0.25
1
Mean of X
Var of X
StDev of X
0.75
0.25
2
0
1.25
0.25
4
0.5
0.707107
6
16.8
11
33.8
4
6
8
xyP(x)
cov(x,y) =
sum xyP(x)-muxmuy
5 P(y)
0.05
0.1
0.1
Mean of Y Var of Y
StDev of Y
0.3
1.2
1.2
0.4
2.4 3.15544E-31
0.3
2.4
1.2
6
2.4 1.549193
9.8
W  a  x  b y = (1)4 + (1)6 = 10
 2W  a 2 2 X  b 2 2Y  2abCov( X , Y )  12 (.5)  12 (2.4)  2(1)(1)(9.8)  22.5