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 =