Discrete Random Variables and Probabilitydistributions

Chapter 5: Discrete Random Variables and Probability Distributions 1

Chapter 5:

Discrete Random Variables and ProbabilityDistributions

5.1Daily computer sales is a discrete random variable that can take on no more than a countable number of values

5.2The 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.

5.3a. Discrete – a countable number

1. Discrete - countable
2. 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
3. Discrete - countable

5.4Discrete random variable – number of plays is countable

5.5Various 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.

5.6Total sales, advertising expenditures, sales of competitors

5.7Discrete – the number of voters supporting a candidate is a countable number of values

5.8Discrete – the number of purchases is a countable number of values

5.9Probability distribution of the number of heads

X-number of heads / P(x)
0 / .5
1 / .5

5.10Probability distribution of number of heads in one toss

X-number of heads / P(x)
0 / .5
1 / .5

5.11Probability distribution of number of heads when three coins are tossed

X-number of heads / P(x)
0 / .125
1 / .375
2 / .375
3 / .125

5.12Various answers

X –# of times missing class / P(x) / F(x)
0 / .65 / .65
1 / .15 / .80
2 / .10 / .90
3 / .09 / .99
4 / .01 / 1.00

5.13a. P(3 ≤ x <) = .20 + .20 + .15 = 0.55

b. P(x > 3) = .20 + .15 + .10 = 0.45

c. P(x ≤ 4) = .05 + .10 + .20+ .20 + .20 = 0.75

d. P(2 < x ≤ 5) = .20 + .20 + .15 = 0.55

5.14a. Cumulative probability function:

X / 0 / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9
P(x) / .10 / .08 / .07 / .15 / .12 / .08 / .10 / .12 / .08 / .10
F(x) / .10 / .18 / .25 / .40 / .52 / .60 / .70 / .82 / .90 / 1.00

1. P(x ≥ 5) = .08 + .10 + .12 + .08 + .10 = .48
2. P(3 ≤ x ≤ 7) = .15 + .12 + .08 + .10 + .12 = .57

5.15a. Draw the probability distribution function

b. Calculate and draw the cumulative probability function

X / P(x) / F(x)
0 / .40 / .40
1 / .60 / 1.00

c. Find the mean of the random variable of x

X / P(x) / XP(x)
0 / .40 / 0
1 / .60 / .60
.60

= .60

d. Find the variance of X

X / P(x) / XP(x) / (x-mu)^2 / (x-mu)^2P(x)
0 / .40 / 0 / .36 / .144
1 / .60 / .60 / .16 / .096
.60 / .240

= .240

5.16a. Probability distribution function

b. Cumulative probability function

c. Find the mean

X / P(x) / XP(x)
0 / .25 / 0
1 / .50 / .50
2 / .25 / .50
1.00

= 1.00

d. Find the variance of X

X / P(x) / XP(x) / (x-mu)^2 / (x-mu)^2P(x)
0 / .25 / 0 / 1.0 / .25
1 / .50 / .50 / 0 / 0
2 / .25 / .50 / 1 / .25
1.00 / .50

= .50

5.17a. Probability distribution function

b. Cumulative probability distribution function

c. Find the mean of the random variable of x

X / P(x) / XP(x)
0 / .50 / 0
1 / .50 / .50
.50

= .50

d. Find the variance of X

X / P(x) / XP(x) / (x-mu)^2 / (x-mu)^2P(x)
0 / .50 / 0 / .25 / .125
1 / .50 / .50 / .25 / .125
.50 / .25

= .25

5.18a. Probability function:

b. Cumulative probability function:

c. = 0 + .36 + 2(.23) + 3(.09) + 4(.04) = 1.25 defects

d. = 1.1675

Excel output:

Returns / P(x) / F(x) / Mean / Variance
0 / 0.28 / 0.28 / 0 / 0.4375
1 / 0.36 / 0.64 / 0.36 / 0.0225
2 / 0.23 / 0.87 / 0.46 / 0.129375
3 / 0.09 / 0.96 / 0.27 / 0.275625
4 / 0.04 / 1.00 / 0.16 / 0.3025
1.00 / 1.25 / 1.1675
S.D. / 1.080509

5.19 a. Probability function:

b. Cumulative probability function:

1. P(x ≥ 3) = .50

d. = 2.45 orders

e. = 1.3592 orders

Excel output:

Orders / P(x) / F(x) / Mean / Variance
0 / 0.10 / 0.10 / 0 / 0.60025
1 / 0.14 / 0.24 / 0.14 / 0.29435
2 / 0.26 / 0.50 / 0.52 / 0.05265
3 / 0.28 / 0.78 / 0.84 / 0.0847
4 / 0.15 / 0.93 / 0.6 / 0.360375
5 / 0.07 / 1.00 / 0.35 / 0.455175
1.00 / 2.45 / 1.8475
S.D. / 1.359228

5.20 a. Probability function

b. Cumulative probability function

c. P(49  x  51) = .70

d. 1 – [P(x < 50)]2 = 1 – .1444 = .8556

e. = 47(.04) + 48(.13) + 49(.21) + 50(.29) + 51(.20) + 52(.10) + 53(.03) = 49.9 clips

= 1.95 = 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

5.21 a. Probability function

b. Cumulative probability function:

c. P(x  4) = .32

d. P(X < 3 (both days)) = (.37)2 = .1369

e. = 2.99 riders = 1.9899 = 1.4106 riders

BusRiders / P(x) / F(x) / Mean / Variance
0 / 0.02 / 0.02 / 0 / 0.17880200
1 / 0.12 / 0.14 / 0.12 / 0.47521200
2 / 0.23 / 0.37 / 0.46 / 0.22542300
3 / 0.31 / 0.68 / 0.93 / 0.00003100
4 / 0.19 / 0.87 / 0.76 / 0.19381900
5 / 0.08 / 0.95 / 0.4 / 0.32320800
6 / 0.03 / 0.98 / 0.18 / 0.27180300
7 / 0.02 / 1.00 / 0.14 / 0.32160200
1.00 / 2.99 / 1.98990000
S.D. / 1.41063815

f. Revenue: r = .50X, E(r) = .50(2.99) = 1.495

= |.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

1. 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

= .22 – (.20)2 = .18

d. = 0(153/190) + (36/190) + 2(1/190) = 38/190 = 0.2 defects

= 40/190 – (.20)2 = .1705

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, …

1. 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 = 1.3259 = 1.1515

Rating / P(x) / F(x) / Mean / Variance
1 / 0.07 / 0.07 / 0.07 / 0.367087
2 / 0.19 / 0.26 / 0.38 / 0.316179
3 / 0.28 / 0.54 / 0.84 / 0.023548
4 / 0.30 / 0.84 / 1.20 / 0.15123
5 / 0.16 / 1.00 / 0.80 / 0.467856
1.00 / 3.29 / 1.3259
S.D. / 1.151477

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

Level of profit for all combinations of S and X:

S\X / 0 / 1 / 2 / 3 / 4 / 5
0 / 0.00 / -0.05 / -0.10 / -0.15 / -0.20 / -0.25
1 / -0.70 / 0.20 / 0.15 / 0.10 / 0.05 / 0.00
2 / -1.40 / -0.50 / 0.40 / 0.35 / 0.30 / 0.25
3 / -2.10 / -1.20 / -0.30 / 0.60 / 0.55 / 0.50
4 / -2.80 / -1.90 / -1.00 / -0.10 / 0.80 / 0.75
5 / -3.50 / -2.60 / -1.70 / -0.80 / 0.10 / 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 byordering one newspaper

5.28a. = 1.82 breakdowns = 1.0276 = 1.0137 breakdowns

Breakdowns / P(x) / F(x) / Mean / Variance / Cost / P(x) / F(x) / Mean / Variance
0 / 0.1 / 0.10 / 0 / 0.33124 / 0 / 0.1 / 0.10 / 0 / 745290
1 / 0.26 / 0.36 / 0.26 / 0.174824 / 1500 / 0.26 / 0.36 / 390 / 393354
2 / 0.42 / 0.78 / 0.84 / 0.013608 / 3000 / 0.42 / 0.78 / 1260 / 30618
3 / 0.16 / 0.94 / 0.48 / 0.222784 / 4500 / 0.16 / 0.94 / 720 / 501264
4 / 0.06 / 1.00 / 0.24 / 0.285144 / 6000 / 0.06 / 1.00 / 360 / 641574
1.00 / 1.82 / 1.0276 / 1.00 / 2730 / 2312100
S.D. / 1.013706 / S.D. / 1520.559

1. Cost: C = 1500X

E(C) = 1500(1.82) = = $2,730

= |1500|(1.0137) = $1,520.559

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.

5.30Mean and variance of a Bernoulli random variable with P=.5

5.31Probability 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 P(X<=x)

0 0.000244

1 0.003174

2 0.019287

3 0.072998

4 0.193848

5 0.387207

6 0.612793

7 0.806152

8 0.927002

9 0.980713

P(x=7) = .806152 - .612793 = .1934

P(x<6) = .3872

5.32Probability 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 P(X<=x)

0 0.006782

1 0.047476

2 0.160836

3 0.355167

4 0.584201

5 0.780516

6 0.906718

7 0.968531

8 0.991711

P(x=7) = .968531 - .906718 = .06181

P(x<6) = .7805

5.33Probability 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 P(X<=x)

0 0.000037

1 0.000524

2 0.003611

3 0.015961

4 0.050952

5 0.125599

6 0.250011

7 0.415893

8 0.595599

9 0.755337

10 0.872479

P(x=9) = .755337 - .595599 = .1597

P(x<7) = P(x≤6) = .250011

5.34Probability 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 P(X<=x)

0 0.000000

1 0.000000

2 0.000000

3 0.000004

4 0.000039

5 0.000269

6 0.001430

7 0.006073

8 0.020968

9 0.059586

10 0.140683

11 0.278304

12 0.465620

13 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

1. P(x  1) = 1 – Px(0) = 1 – .2373 = .7627
2. P(x  3) = 1 – P(x ≤ 2) = 1 - .8965 = .1035

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

1. 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

1. P(x = 5) = P(x ≤ 5) – Px ≤ 4) = 1.00 - .9898 = .0102
2. P(x  3) = P(x ≤ 5) – P(x ≤ 2) = 1.000 - .6826 = .3174

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

1. P(x  2) = .5248
2. 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.
3. E(X) = = 1 + np = 1 + 4(.4) = 2.6 games

5.41Find 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

1. P(x  2) = P(X ≤ 4) – P(X ≤ 1) = 1.000 - .4752 = .5248

b. E(X) = np = 4(.4) = 1.6

5.43 a E(X) = 50(.15) = 7.5

b. Let Z = 250X

E(Z) = 250(7.5) = $1,875

= |250|(2.5249) = $631

5.44 a. E(X) = 2000(.032) = 64

b. Let Z = 10X

E(Z) = 10(64) = $640

= |10|(7.871) = $78.71

5.45

= 2.0 sales OR = np = 5(.4) = 2.0 sales

5.46a. E(X) = = np = 620(.78) = 483.6, = = 10.3146

1. Let Z = 2X

E(Z) = 2(483.6) = $967.20

= |2|(10.314) = $20.6292

5.47a. 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

5.48 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

5.49P(Supplier1|x=1) = .916

5.50

Probability Density Function

Hypergeometric with N = 50, M = 25, and n = 12

x P(X=x)

0 0.000043

1 0.000918

2 0.008078

3 0.038706

4 0.112702

5 0.210376

P(x=5) = .210376

5.51

Probability Density Function

Hypergeometric with N = 60, M = 25, and n = 14

x P(X=x)

0 0.000134

1 0.002128

2 0.014432

3 0.055323

4 0.133881

5 0.216269

6 0.240299

7 0.186354

P(x=7) = .186354

5.52

Probability Density Function

Hypergeometric with N = 80, M = 42, and n = 20

x P(X=x)

0 0.000000

1 0.000000

2 0.000008

3 0.000093

4 0.000704

5 0.003723

6 0.014348

7 0.041322

8 0.090392

9 0.151769

P(x=9) = .151769

5.53

Probability Density Function

Hypergeometric with N = 40, M = 25, and n = 5

x P(X=x)

0 0.004564

1 0.051861

2 0.207444

3 0.367017

P(x=3) = .367017

5.54

Probability Density Function

Hypergeometric with N = 400, M = 200, and n = 15

x P(X=x)

0 0.000023

1 0.000375

2 0.002792

3 0.012743

4 0.039848

5 0.090434

6 0.153879

7 0.199906

8 0.199906

P(x=8) = .1999

5.55a. 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

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

5.59

Probability Density Function

Poisson with mean = 3.5

x P(X=x)

0 0.030197

1 0.105691

2 0.184959

3 0.215785

4 0.188812

5 0.132169

6 0.077098

7 0.038549

P(x=7) = .038549

5.60

Probability Density Function

Poisson with mean = 2.5

x P(X=x)

0 0.082085

1 0.205212

2 0.256516

3 0.213763

4 0.133602

P(x=4) = .1336

5.61

Cumulative Distribution Function

Poisson with mean = 4.5

x P(X<=x)

0 0.011109

1 0.061099

2 0.173578

3 0.342296

4 0.532104

5 0.702930

6 0.831051

7 0.913414

8 0.959743

P(x>7) = 1-(Px≤ 7) = 1 - .913414 = .086586

5.62

Cumulative Distribution Function

Poisson with mean = 3.5

x P(X<=x)

0 0.030197

1 0.135888

2 0.320847

3 0.536633

4 0.725445

5 0.857614

6 0.934712

P(x<6) = .857614

5.63

Cumulative Distribution Function

Poisson with mean = 8

x P(X<=x)

0 0.000335

1 0.003019

2 0.013754

3 0.042380

4 0.099632

5 0.191236

6 0.313374

7 0.452961

8 0.592547

9 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

5.65a. 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

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

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.

5.72Two 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.

Cumulative Distribution Function

Poisson with mean = 1

x P(X<=x)

0 0.36788

1 0.73576

2 0.91970

3 0.98101

4 0.99634

5 0.99941

6 0.99992

7 0.99999

8 1.00000

9 1.00000

10 1.00000

Cumulative Distribution Function

Binomial with n = 10 and p = 0.1

x P(X<=x)

0 0.34868

1 0.73610

2 0.92981

3 0.98720

4 0.99837

5 0.99985

6 0.99999

7 1.00000

8 1.00000

9 1.00000

10 1.00000

5.73a. Compute marginal probability distributions for X and Y

Exercise_5.73 / X_5.73
Y_5.73 / 1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.25 / 0.25 / 0.5 / 0 / 0.125
1 / 0.25 / 0.25 / 0.5 / 0.5 / 0.125
P(x) / 0.5 / 0.5 / 1 / 0.5 / 0.25 / 0.5
Mean of X / 0.5 / 1 / 1.5
Var of X / 0.125 / 0.125 / 0.25
StDev of X / 0.5
xyP(x) / 0.25 / 0.5 / 0.75
Cov(x,y) =
sum xyP(x)-muxmuy / 0

b. Compute the covariance and correlation for X and Y

= .75 – (1.5)(.5) = 0.0

= 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.

5.74a. Compute marginal probability distributions for X and Y

Exercise_5.74 / X_5.74
Y_5.74 / 1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.2 / 0.25 / 0.45 / 0 / 0.136125
1 / 0.3 / 0.25 / 0.55 / 0.55 / 0.111375
P(x) / 0.5 / 0.5 / 1 / 0.55 / 0.2475 / 0.497494
Mean of X / 0.5 / 1 / 1.5
Var of X / 0.125 / 0.125 / 0.25
StDev of X / 0.5
xyP(x) / 0.3 / 0.5 / 0.8
Cov(x,y) =
sum xyP(x)-muxmuy / -0.025

b. Compute the covariance and correlation for X and Y

= .80 – (1.5)(.55) = -.025

= -.025/(.5)(.497494) = -.1005

5.75 a. Compute marginal probability distributions for X and Y

Exercise_5.75 / X_5.75
Y_5.75 / 1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.25 / 0.25 / 0.5 / 0 / 0.125
1 / 0.25 / 0.25 / 0.5 / 0.5 / 0.125
P(x) / 0.5 / 0.5 / 1 / 0.5 / 0.25 / 0.5
Mean of X / 0.5 / 1 / 1.5
Var of X / 0.125 / 0.125 / 0.25
StDev of X / 0.5
xyP(x) / 0.25 / 0.5 / 0.75
Cov(x,y) =
sum xyP(x)-muxmuy / 0

b. Compute the covariance and correlation for X and Y

= .75 – (1.5)(.5) = 0.0

= 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

= (1)1.5 + (1).5 = 2.0

5.76 a. Compute marginal probability distributions for X and Y

Exercise_5.76 / X_5.76
Y_5.76 / 1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.3 / 0.2 / 0.5 / 0 / 0.125
1 / 0.25 / 0.25 / 0.5 / 0.5 / 0.125
P(x) / 0.55 / 0.45 / 1 / 0.5 / 0.25 / 0.5
Mean of X / 0.55 / 0.9 / 1.45
Var of X / 0.55 / 1.8 / 2.35
StDev of X / 1.532971
xyP(x) / 0.25 / 0.5 / 0.75
Cov(x,y) =
sum xyP(x)-muxmuy / 0.025

b. Compute the covariance and correlation for X and Y

= .75 – (1.45)(.5) = 0.025

= 0.025/(1.53297)(.5) = 0.0326

c. Compute the mean and variance for the linear function W = 2X + Y

= (2)1.45 + (1).5 = 3.4

5.77 a. Compute marginal probability distributions for X and Y

Exercise_5.77 / X_5.77
Y_5.77 / 1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.7 / 0 / 0.7 / 0 / 0.063
1 / 0 / 0.3 / 0.3 / 0.3 / 0.147
P(x) / 0.7 / 0.3 / 1 / 0.3 / 0.21 / 0.458258
Mean of X / 0.7 / 0.6 / 1.3
Var of X / 0.063 / 0.147 / 0.21
StDev of X / 0.458258
xyP(x) / 0 / 0.6 / 0.6
Cov(x,y) =
sum xyP(x)-muxmuy / 0.21

b. Compute the covariance and correlation for X and Y

= .60 – (1.3)(.3) = 0.21

= 0.21/(.458258)(.458258) = 1.00

c. Compute the mean and variance for the linear function W = 3X + 4Y

= (3)1.3 + (4).3 = 5.1

5.78 a. Compute the marginal probability distributions for X and Y

1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.25 / 0.25 / 0.5 / 0 / 0.125
1 / 0.25 / 0.25 / 0.5 / 0.5 / 0.125
P(x) / 0.5 / 0.5 / 1 / 0.5 / 0.25 / 0.5
Mean of X / 0.5 / 1 / 1.5
Var of X / 0.125 / 0.125 / 0.25
StDev of X / 0.5
xyP(x) / 0.25 / 0.5 / 0.75
Cov(x,y) =
sum xyP(x)-muxmuy / 0

b. Compute the covariance and correlation for X and Y

= .75 – (1.5)(.5) = 0.0

= 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

= (1)1.5 + (-1).5 = 1.0

5.79a. Compute the marginal probability distributions for X and Y.

1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.3 / 0.2 / 0.5 / 0 / 0.125
1 / 0.25 / 0.25 / 0.5 / 0.5 / 0.125
P(x) / 0.55 / 0.45 / 1 / 0.5 / 0.25 / 0.5
Mean of X / 0.55 / 0.9 / 1.45
Var of X / 0.55 / 1.8 / 2.35
StDev of X / 1.532971
xyP(x) / 0.25 / 0.5 / 0.75
Cov(x,y) =
sum xyP(x)-muxmuy / 0.025

b. Compute the covariance and correlation for X and Y

= .75 – (1.45)(.5) = 0.025

= 0.025/(1.53297)(.5) = 0.0326

c. Compute the mean and variance for the linear function W = 2X - Y

= (2)1.45 + (-1).5 = 2.4

5.80a. Compute the marginal probability distributions for X and Y.

Exercise_5.80 / X_5.80
Y_5.80 / 1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0 / 0.6 / 0.6 / 0 / 0.096
1 / 0.4 / 0 / 0.4 / 0.4 / 0.144
P(x) / 0.4 / 0.6 / 1 / 0.4 / 0.24 / 0.489898
Mean of X / 0.4 / 1.2 / 1.6
Var of X / 0.144 / 0.096 / 0.24
StDev of X / 0.489898
xyP(x) / 0.4 / 0 / 0.4
Cov(x,y) =
sum xyP(x)-muxmuy / -0.24

b. Compute the covariance and correlation for X and Y

= .40 – (1.6)(.4) = -0.24

= -0.24/(.489898)(.489898) = -1.00

c. Compute the mean and variance for the linear function W = 2X - 4Y

= (2)1.6 + (-4).4 = 1.6

5.81a. Compute marginal probability distributions for X and Y

1 / 2 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.7 / 0 / 0.7 / 0 / 0.063
1 / 0 / 0.3 / 0.3 / 0.3 / 0.147
P(x) / 0.7 / 0.3 / 1 / 0.3 / 0.21 / 0.458258
Mean of X / 0.7 / 0.6 / 1.3
Var of X / 0.063 / 0.147 / 0.21
StDev of X / 0.458258
xyP(x) / 0 / 0.6 / 0.6
Cov(x,y) =
sum xyP(x)-muxmuy / 0.21

b. Compute the covariance and correlation for X and Y

= .60 – (1.3)(.3) = 0.21

= 0.21/(.458258)(.458258) = 1.00

c. Compute the mean and variance for the linear function W = 10X - 8Y

= (10)1.3 + (-8).3 = 10.6

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

= 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

= 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.

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

= .3399

e. No, because

5.83a. 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

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

= .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

Y Return
X Return / 0 / 1 / 2 / P(x) / Mean of X / Var of X / StDev of X
3 / 0.09 / 0.14 / 0.07 / 0.3 / 0.9 / 0.26508
4 / 0.07 / 0.23 / 0.16 / 0.46 / 1.84 / 0.001656
5 / 0.03 / 0.1 / 0.11 / 0.24 / 1.2 / 0.269664
P(y) / 0.19 / 0.47 / 0.34 / 3.94 / 0.5364 / 0.732393
Mean of Y / 0 / 0.47 / 0.68 / 1.15
Var of Y / 0.251275 / 0.010575 / 0.24565 / 0.5075
StDev of Y / 0.4956309
xyP(x) / 0 / 1.84 / 2.8 / 4.64
sum xyP(x)*muxmuy / 0.109

5.84 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

1. No, because Px,y(3,4) = .08 ≠ .0468 = Px(3)Py(4)

5.85a. 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

The covariance indicates that there is a positive association between brand watchers of a late-night talk show and brand name recognition.

Y Identify
X Watch / 0 / 1 / P(x) / Mean of X / Var of X / StDev of X
0 / 0.54 / 0.3 / 0.84 / 0 / 0.021504
1 / 0.01 / 0.15 / 0.16 / 0.16 / 0.112896
P(y) / 0.55 / 0.45 / 0.16 / 0.1344 / 0.366606056
Mean of Y / 0 / 0.45 / 0.45
Var of Y / 0.111375 / 0.136125 / 0.2475
StDev of Y / 0.49749372
xyP(x) / 0 / 0.15 / 0.15
Sum xyP(x)*muxmuy / 0.078

5.86a.

Y/X / 0 / 1 / Total
0 / .704 / .168 / .872
1 / .096 / .032 / .128
Total / .80 / .20 / 1.00

1. PY|X(y|0) = .88; .12
2. Px(0) = .80

Px(1) = .20

Py(0) = .872

Py(1) = .128

d. E(XY) = .032;

,

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.

5.87Because of independence, the joint probabilities are the products of the marginal probabilities, so P(0,0)=.0216, and so on.

Y Service
X Food / 0 / 1 / 2 / 3 / P(x) / Mean of X / Var of X / StDev of X
0 / 0.0216 / 0.0456 / 0.0408 / 0.012 / 0.12 / 0 / 0.322752
1 / 0.0522 / 0.1102 / 0.0986 / 0.029 / 0.29 / 0.29 / 0.118784
2 / 0.0756 / 0.1596 / 0.1428 / 0.042 / 0.42 / 0.84 / 0.054432
3 / 0.0306 / 0.0646 / 0.0578 / 0.017 / 0.17 / 0.51 / 0.314432
P(y) / 0.18 / 0.38 / 0.34 / 0.1 / 1 / 1.64 / 0.8104 / 0.900222
Mean of Y / 0 / 0.38 / 0.68 / 0.3 / 1.36
Var of Y / 0.332928 / 0.049248 / 0.139264 / 0.26896 / 0.7904
StDev of Y / 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 / P(x) / Mean of X / Var of X / StDev of X
0 / 0.0144 / 0.0208 / 0.0288 / 0.0104 / 0.0056 / 0.08 / 0 / 0.453152
1 / 0.0288 / 0.0416 / 0.0576 / 0.0208 / 0.0112 / 0.16 / 0.16 / 0.304704
2 / 0.0504 / 0.0728 / 0.1008 / 0.0364 / 0.0196 / 0.28 / 0.56 / 0.040432
3 / 0.0576 / 0.0832 / 0.1152 / 0.0416 / 0.0224 / 0.32 / 0.96 / 0.123008
4 / 0.018 / 0.026 / 0.036 / 0.013 / 0.007 / 0.1 / 0.4 / 0.26244
5 / 0.0108 / 0.0156 / 0.0216 / 0.0078 / 0.0042 / 0.06 / 0.3 / 0.411864
P(y) / 0.18 / 0.26 / 0.36 / 0.13 / 0.07 / 1 / 2.38 / 1.5956 / 1.263171
Mean of Y / 0 / 0.26 / 0.72 / 0.39 / 0.28 / 1.65
Var of Y / 0.49005 / 0.10985 / 0.0441 / 0.23693 / 0.38658 / 1.2675
StDev of Y / 1.12583302

= 5(2.38) + 10(1.65) = 28.4

= 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 / P(x) / F(x) / Mean / Variance
1 / 0.05 / 0.05 / 0.05 / 0.242
2 / 0.2 / 0.25 / 0.4 / 0.288
3 / 0.35 / 0.60 / 1.05 / 0.014
4 / 0.3 / 0.90 / 1.2 / 0.192
5 / 0.1 / 1.00 / 0.5 / 0.324
Ex 5.56 / 1.00 / 3.2 / 1.06
S.D. / 1.029563

1. P(x < 3) = .05+ .20 = .25
2. E(X) = 3.2
3. = 1.029563
4. Cost = $20,000 + $2,000X = E(Cost) = $26,400,

standard deviation = ($2,000)(1.029563) = $2,059.13

1. 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 / P(x) / F(x) / Mean / Variance
0 / 0.1 / 0.10 / 0 / 0.48841
1 / 0.2 / 0.30 / 0.2 / 0.29282
2 / 0.35 / 0.65 / 0.7 / 0.015435
3 / 0.16 / 0.81 / 0.48 / 0.099856
4 / 0.12 / 0.93 / 0.48 / 0.384492
5 / 0.07 / 1.00 / 0.35 / 0.544887
Ex 5.57 / 1.00 / 2.21 / 1.8259
S.D. / 1.351259

1. E(X) = 2.21 cars sold
2. Standard deviation = 1.3513 cars
3. Mean Salary = $250 + $300 (2.21) = $913. Standard deviation of salary = $300(1.3513) = $405.39
4. To earn a salary of $1,000 or more, the salesperson must sell at least 3 cars.

P(X  3) = .16 + .12 + .07 = .35

5.93a.  = np = 9(.25) = 2.25

b) = = 1.299

c) (i) E(X) = 1 + 2.23 = 3.25, (ii) = 1.299

5.94 a. Positive covariance: Consumption expenditures & Disposable income

1. Negative covariance: Price of cars and the number of cars sold
2. 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. , vehicles

5.96

X Years
Y Visits / 1 / 2 / 3 / 4 / P(y) / Mean of Y / Var of Y / StDev of Y
0 / 0.07 / 0.05 / 0.03 / 0.02 / 0.17 / 0 / 0.2057
1 / 0.13 / 0.11 / 0.17 / 0.15 / 0.56 / 0.56 / 0.0056
2 / 0.04 / 0.04 / 0.09 / 0.1 / 0.27 / 0.54 / 0.2187
P(x) / 0.24 / 0.2 / 0.29 / 0.27 / 1 / 1.1 / 0.43 / 0.6557439
Mean of X / 0.24 / 0.4 / 0.87 / 1.08 / 2.59
Var of X / 0.606744 / 0.0696 / 0.048749 / 0.5368 / 1.2619
StDev of X / 1.12334
xyP(x) / 0.21 / 0.38 / 1.05 / 1.4 / 3.04
sum xyP(x)*muxmuy / 0.191

1. Py(0) = .07+.05+.03+.02 = .17
2. E(X) =

E(Y) =

1. 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.

5.97Assume that the shots are independent of each other

a.P(x  2) = 1 – P(x  1) = 1 – [( ] = 0.767

b. P(x=3) = = 0.2765

c. = (6)(.4) = 2.4, = 1.2

d. Mean of total points scored = 3() = 3(2.4) = 7.2, Std dev = 3 = 3(1.2)=3.6

5.98a. P(x=3) = = .3369

1. P(x  3) = P(3)+P(4)+P(5) = .3369 + (5)(.55)4(.45) + (1)(.55)5(1) = .5931
2. = (80)(.55) = 44 will graduate in 4 years. The proportion is 44/80 = .55. = 4.4497. The proportion is 4.4497/80 = .05562

5.99a. This is a binomial probability (assuming independence) with a p=.6 and n=7. Then the P(A wins) = P(X ≥ 4) = = 0.71021

b. = 0.27648

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, (ii) = 0.48

5.100To 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) = = .16683

5.101Find P(X  2) = 1 - = .16202

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

5.1031 – e-6.5 - e-6.5(6.5) - e-6.5 (6.5)2/2! = 0.95696

5.104P(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

5.105compute the mean and variance for the total value of the stock portfolio

Exercise_5.105 / X_5.105
Y_5.105 / 40 / 50 / 60 / 70 / P(y) / Mean of Y / Var of Y / StDev of Y
45 / 0 / 0 / 0.05 / 0.2 / 0.25 / 11.25 / 17.01563
50 / 0.05 / 0 / 0.05 / 0.1 / 0.2 / 10 / 2.1125
55 / 0.1 / 0.05 / 0 / 0.05 / 0.2 / 11 / 0.6125
60 / 0.2 / 0.1 / 0.05 / 0 / 0.35 / 21 / 15.94688
P(x) / 0.35 / 0.15 / 0.15 / 0.35 / 1 / 53.25 / 35.6875 / 5.973902
Mean of X / 14 / 7.5 / 9 / 24.5 / 55
Var of X / 78.75 / 3.75 / 3.75 / 78.75 / 165
StDev of X / 12.84523
xyP(x) / 900 / 437.5 / 615 / 1522.5 / 3475
cov(x,y) =
sum xyP(x)-muxmuy / 546.25

Compute the mean and variance for the linear function W = X + Y

= (1)55 + (1)53.25 = 108.25

5.106 Compute the mean and variance

Exercise_5.106 / X_5.106
Y_5.106 / 3 / 4 / 5 / P(y) / Mean of Y / Var of Y / StDev of Y
4 / 0.1 / 0.15 / 0.05 / 0.3 / 1.2 / 1.2
6 / 0.1 / 0.2 / 0.1 / 0.4 / 2.4 / 3.15544E-31
8 / 0.05 / 0.15 / 0.1 / 0.3 / 2.4 / 1.2
P(x) / 0.25 / 0.5 / 0.25 / 1 / 6 / 2.4 / 1.549193
Mean of X / 0.75 / 2 / 1.25 / 4
Var of X / 0.25 / 0 / 0.25 / 0.5
StDev of X / 0.707107
xyP(x) / 6 / 16.8 / 11 / 33.8
cov(x,y) =
sum xyP(x)-muxmuy / 9.8

= (1)4 + (1)6 = 10