Discrete Probability Distributions

Chapter 5

Discrete Probability Distributions

Learning Objectives

1. Understand the concepts of a random variable and a probability distribution.

2. Be able to distinguish between discrete and continuous random variables.

3. Be able to compute and interpret the expected value, variance, and standard deviation for a discrete random variable.

4. Be able to compute and work with probabilities involving a binomial probability distribution.

5. Be able to compute and work with probabilities involving a Poisson probability distribution.

6. Know when and how to use the hypergeometric probability distribution.

Solutions:

1. a. Head, Head (H,H)

Head, Tail (H,T)

Tail, Head (T,H)

Tail, Tail (T,T)

b. x = number of heads on two coin tosses

Outcome / Values of x
(H,H) / 2
(H,T) / 1
(T,H) / 1
(T,T) / 0

d. Discrete. It may assume 3 values: 0, 1, and 2.

2. a. Let x = time (in minutes) to assemble the product.

b. It may assume any positive value: x > 0.

c. Continuous

3. Let Y = position is offered

N = position is not offered

a. S = {(Y,Y,Y), (Y,Y,N), (Y,N,Y), (Y,N,N), (N,Y,Y), (N,Y,N), (N,N,Y), (N,N,N)}

b. Let N = number of offers made; N is a discrete random variable.

Experimental Outcome / (Y,Y,Y) / (Y,Y,N) / (Y,N,Y) / (Y,N,N) / (N,Y,Y) / (N,Y,N) / (N,N,Y) / (N,N,N)
Value of N / 3 / 2 / 2 / 1 / 2 / 1 / 1 / 0

4. x = 0, 1, 2, . . ., 12.

5. a. S = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}

Experimental Outcome / (1,1) / (1,2) / (1,3) / (2,1) / (2,2) / (2,3)
Number of Steps Required / 2 / 3 / 4 / 3 / 4 / 5

6. a. values: 0,1,2,...,20

discrete

b. values: 0,1,2,...

discrete

c. values: 0,1,2,...,50

discrete

d. values: 0 £ x £ 8

continuous

e. values: x > 0

continuous

7. a. f (x) ³ 0 for all values of x.

S f (x) = 1 Therefore, it is a proper probability distribution.

b. Probability x = 30 is f (30) = .25

c. Probability x £ 25 is f (20) + f (25) = .20 + .15 = .35

d. Probability x > 30 is f (35) = .40

8. a.

x / f (x)
1 / 3/20 = .15
2 / 5/20 = .25
3 / 8/20 = .40
4 / 4/20 = .20
Total 1.00

c. f (x) ³ 0 for x = 1,2,3,4.

S f (x) = 1

9. a.

Age / Number of Children / f(x)
6 / 37,369 / 0.018
7 / 87,436 / 0.043
8 / 160,840 / 0.080
9 / 239,719 / 0.119
10 / 286,719 / 0.142
11 / 306,533 / 0.152
12 / 310,787 / 0.154
13 / 302,604 / 0.150
14 / 289,168 / 0.143
2,021,175 / 1.001

c. f(x) ³ 0 for every x

S f(x) = 1

Note: S f(x) = 1.001 in part (a); difference from 1 is due to rounding values of f(x).

10. a.

x / f(x)
1 / 0.05
2 / 0.09
3 / 0.03
4 / 0.42
5 / 0.41
1.00

x / f(x)
1 / 0.04
2 / 0.10
3 / 0.12
4 / 0.46
5 / 0.28
1.00

c. P(4 or 5) = f (4) + f (5) = 0.42 + 0.41 = 0.83

d. Probability of very satisfied: 0.28

e. Senior executives appear to be more satisfied than middle managers. 83% of senior executives have a score of 4 or 5 with 41% reporting a 5. Only 28% of middle managers report being very satisfied.

11. a.

Duration of Call
x / f(x)
1 / 0.25
2 / 0.25
3 / 0.25
4 / 0.25
1.00

c. f (x) ³ 0 and f (1) + f (2) + f (3) + f (4) = 0.25 + 0.25 + 0.25 + 0.25 = 1.00

d. f (3) = 0.25

e. P(overtime) = f (3) + f (4) = 0.25 + 0.25 = 0.50

12. a. Yes; f (x) ³ 0 for all x and S f (x) = .15 + .20 + .30 + .25 + .10 = 1

b. P(1200 or less) = f (1000) + f (1100) + f (1200)

= .15 + .20 + .30 = .65

13. a. Yes, since f (x) ³ 0 for x = 1,2,3 and S f (x) = f (1) + f (2) + f (3) = 1/6 + 2/6 + 3/6 = 1

b. f (2) = 2/6 = .333

c. f (2) + f (3) = 2/6 + 3/6 = .833

14. a. f (200) = 1 - f (-100) - f (0) - f (50) - f (100) - f (150)

= 1 - .95 = .05

This is the probability MRA will have a $200,000 profit.

b. P(Profit) = f (50) + f (100) + f (150) + f (200)

= .30 + .25 + .10 + .05 = .70

c. P(at least 100) = f (100) + f (150) + f (200)

= .25 + .10 +.05 = .40

15. a.

x / f (x) / x f (x)
3 / .25 / .75
6 / .50 / 3.00
9 / .25 / 2.25
1.00 / 6.00

E(x) = m = 6

x / x - m / (x - m)2 / f (x) / (x - m)2 f (x)
3 / -3 / 9 / .25 / 2.25
6 / 0 / 0 / .50 / 0.00
9 / 3 / 9 / .25 / 2.25
4.50

Var(x) = s2 = 4.5

c. s = = 2.12

16. a.

y / f (y) / y f (y)
2 / .2 / .4
4 / .3 / 1.2
7 / .4 / 2.8
8 / .1 / .8
1.0 / 5.2

E(y) = m = 5.2

y / y - m / (y - m)2 / f (y) / (y - m)2 f (y)
2 / -3.20 / 10.24 / .20 / 2.048
4 / -1.20 / 1.44 / .30 / .432
7 / 1.80 / 3.24 / .40 / 1.296
8 / 2.80 / 7.84 / .10 / .784
4.560

17. a/b.

x / f (x) / x f (x) / x - m / (x - m)2 / (x - m)2 f (x)
0 / .10 / .00 / -2.45 / 6.0025 / .600250
1 / .15 / .15 / -1.45 / 2.1025 / .315375
2 / .30 / .60 / - .45 / .2025 / .060750
3 / .20 / .60 / .55 / .3025 / .060500
4 / .15 / .60 / 1.55 / 2.4025 / .360375
5 / .10 / .50 / 2.55 / 6.5025 / .650250
2.45 / 2.047500

E(x) = m = 2.45

s2 = 2.05

s = 1.43

18. a/b.

x / f (x) / xf (x) / x - m / (x - m)2 / (x - m)2 f (x)
0 / 0.04 / 0.00 / -1.84 / 3.39 / 0.12
1 / 0.34 / 0.34 / -0.84 / 0.71 / 0.24
2 / 0.41 / 0.82 / 0.16 / 0.02 / 0.01
3 / 0.18 / 0.53 / 1.16 / 1.34 / 0.24
4 / 0.04 / 0.15 / 2.16 / 4.66 / 0.17
Total / 1.00 / 1.84 / 0.79
E(x) / Var(x)

c/d.

y / f (y) / yf (y) / y - m / (y - m)2 / (y - m)2 f (y)
0 / 0.00 / 0.00 / -2.93 / 8.58 / 0.01
1 / 0.03 / 0.03 / -1.93 / 3.72 / 0.12
2 / 0.23 / 0.45 / -0.93 / 0.86 / 0.20
3 / 0.52 / 1.55 / 0.07 / 0.01 / 0.00
4 / 0.22 / 0.90 / 1.07 / 1.15 / 0.26
Total / 1.00 / 2.93 / 0.59
E(y) / Var(y)

e. The number of bedrooms in owner-occupied houses is greater than in renter-occupied houses. The expected number of bedrooms is 1.09 = 2.93 - 1.84 greater. And, the variability in the number of bedrooms is less for the owner-occupied houses.

19. a. E(x) = S x f (x) = 0 (.56) + 2 (.44) = .88

b. E(x) = S x f (x) = 0 (.66) + 3 (.34) = 1.02

c. The expected value of a 3 - point shot is higher. So, if these probabilities hold up, the team will make more points in the long run with the 3 - point shot.

20. a.

x / f(x) / xf(x)
0 / .85 / 0
500 / .04 / 20
1000 / .04 / 40
3000 / .03 / 90
5000 / .02 / 100
8000 / .01 / 80
10000 / .01 / 100
Total / 1.00 / 430

The expected value of the insurance claim is $430. If the company charges $430 for this type of collision coverage, it would break even.

b. From the point of view of the policyholder, the expected gain is as follows:

Expected Gain = Expected claim payout – Cost of insurance coverage

= $430 - $520 = -$90

The policyholder is concerned that an accident will result in a big repair bill if there is no insurance coverage. So even though the policyholder has an expected annual loss of $90, the insurance is protecting against a large loss.

21. a. E(x) = S x f (x) = 0.05(1) + 0.09(2) + 0.03(3) + 0.42(4) + 0.41(5) = 4.05

b. E(x) = S x f (x) = 0.04(1) + 0.10(2) + 0.12(3) + 0.46(4) + 0.28(5) = 3.84

c. Executives: s2 = S (x - m)2 f(x) = 1.25

Middle Managers: s2 = S (x - m)2 f(x) = 1.13

d. Executives: s = 1.12

Middle Managers: s = 1.07

e. The senior executives have a higher average score: 4.05 vs. 3.84 for the middle managers. The executives also have a slightly higher standard deviation.

22. a. E(x) = S x f (x) = 300 (.20) + 400 (.30) + 500 (.35) + 600 (.15) = 445

The monthly order quantity should be 445 units.

b. Cost: 445 @ $50 = $22,250

Revenue: 300 @ $70 = 21,000

$ 1,250 Loss

23. a. Rent Controlled: E(x) = 1(.61) + 2(.27) + 3(.07) + 4(.04) + 5(.01) = 1.57

Rent Stabilized: E(x) = 1(.41) + 2(.30) + 3(.14) + 4(.11) + 5(.03) + 6(.01) = 2.08

b. Rent Controlled:

Var(x) = (-.57)2.61+ (.43)2.27+ (1.43)2.07+ (2.43)2.04+ (3.43)2.01= .75

Rent Stabilized:

Var(x) = (-1.08)2.41+ (-.08)2.30+ (.92)2.14+ (1.92)2.11+ (2.92)2.03+ (3.92)2.01= 1.41

c. From the expected values in part (a), it is clear that the expected number of persons living in rent stabilized units is greater than the number of persons living in rent controlled units. For example, comparing a building that contained 10 rent controlled units to a building that contained 10 rent stabilized units, the expected number of persons living in the rent controlled building would be 1.57(10) = 15.7 or approximately 16. For the rent stabilized building, the expected number of persons is approximately 21. There is also more variability in the number of persons living in rent stabilized units.

24. a. Medium E(x) = S x f (x)

= 50 (.20) + 150 (.50) + 200 (.30) = 145

Large: E(x) = S x f (x)

= 0 (.20) + 100 (.50) + 300 (.30) = 140

Medium preferred.

b. Medium

x / f (x) / x - m / (x - m)2 / (x - m)2 f (x)
50 / .20 / -95 / 9025 / 1805.0
150 / .50 / 5 / 25 / 12.5
200 / .30 / 55 / 3025 / 907.5

s2 = 2725.0

Large

y / f (y) / y - m / (y - m)2 / (y - m)2 f (y)
0 / .20 / -140 / 19600 / 3920
100 / .50 / -40 / 1600 / 800
300 / .30 / 160 / 25600 / 7680

s2 = 12,400

Medium preferred due to less variance.

25. a.

e. P(x ³ 1) = f (1) + f (2) = .48 + .16 = .64

f. E(x) = n p = 2 (.4) = .8

Var(x) = n p (1 - p) = 2 (.4) (.6) = .48

s = = .6928

26. a. f (0) = .3487

b. f (2) = .1937

c. P(x £ 2) = f (0) + f (1) + f (2) = .3487 + .3874 + .1937 = .9298

d. P(x ³ 1) = 1 - f (0) = 1 - .3487 = .6513

e. E(x) = n p = 10 (.1) = 1

f. Var(x) = n p (1 - p) = 10 (.1) (.9) = .9

s = = .9487

27. a. f (12) = .1144

b. f (16) = .1304

c. P(x ³ 16) = f (16) + f (17) + f (18) + f (19) + f (20)

= .1304 + .0716 + .0278 + .0068 + .0008 = .2374

d. P(x £ 15) = 1 - P (x ³ 16) = 1 - .2374 = .7626

e. E(x) = n p = 20(.7) = 14

f. Var(x) = n p (1 - p) = 20 (.7) (.3) = 4.2

s = = 2.0494

28. a.

b. P(at least 2) = 1 - f(0) - f(1)

= 1 - .2084 - .3735 = .4181

29. a.

b. P(x 3) = 1 - f (0) - f (1) - f (2)