Chapter 5
Discrete Probability Distributions

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

c.
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), (N,Y,Y), (Y,N,N), (N,Y,N), (N,N,Y), (N,N,N)}

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

c.

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

4. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

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

b.

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)  0 for all values of x.

 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

b.

c. f (x)  0 for x = 1,2,3,4.

 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
1.000

b. c. f(x)  0 for every x

 f(x) = 1 10. a. x f(x)
1
0.05
2
0.09
3
0.03
4
0.42
5
0.41

1.00

b. 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 b.

c. f (x)  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)  0.  f (x) = 1

b. f (500,000) + f (600,000) = .10 + .05 = .15

c. f (100,000) = .10

13. a. Yes, since f (x)  0 for x = 1,2,3 and  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) =  = 6

b. x x - 
(x - )2 f (x)
(x - )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) = 2 = 4.5

c.  = = 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) =  = 5.2

b. y y - 
(y - )2 f (y)
(y - )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. Total Student =