1. A mail-order computer business has six telephone lines. Let X ...

= 1 × 0.081920 + 2 × 0.015360 + 3 × 0.001536 + 4 × 0.000064
= 0.117504
(c) Suppose the probability distribution of the number of reservations
made is given in the accompanying table.
Number of reservations 3 4 5 6
Probability .1 .2 .3 .4
Let X denote the number of passengers on a randomly selected
trip. Obtain the probability mass function of X.
Ans
Define the r.v. R to be the no. of reservations. First consider the
case, X = 0, this can happen even if R = 3, 4, 5, 6. Therefore
P [X = 0] = P [X = 0, R = 3]+P [X = 0, R = 4]+P [X = 0, R = 5]+P [X = 0, R = 6]
3
= (0.1) 0.8
0
0 (1 − 0.8) 3−0
4
+ (0.2) 0.8
0
0 (1 − 0.8) 4−0
5
+(0.3) 0.8
0
0 (1 − 0.8) 5−0
6
+ (0.4) 0.8
0
0 (1 − 0.8) 6−0
= 0.0012416
P [X = 1] = P [X = 1, R = 3]+P [X = 1, R = 4]+P [X = 1, R = 5]+P [X = 1, R = 6]
3
= (0.1) 0.8
1
1 (1 − 0.8) 3−1
4
+ (0.2) 0.8
1
1 (1 − 0.8) 4−1
5
+(0.3) 0.8
1
1 (1 − 0.8) 5−1
6
+ (0.4) 0.8
1
1 (1 − 0.8) 6−1
0.0172544
P [X = 2] = P [X = 2, R = 3]+P [X = 2, R = 4]+P [X = 2, R = 5]+P [X = 2, R = 6]
= 0.090624
P [X = 3] = P [X = 3, R = 3]+P [X = 3, R = 4]+P [X = 3, R = 5]+P [X = 3, R = 6]
= 0.227328
P [X = 4] = 1−P [X = 0]−P [X = 1]−P [X = 2]−P [X = 3] = 0.663552
9. The article ”Modeling Sediment and Water Column Interactions for
Hydrophobic Pollutants” (Water Research, 1984, pp. 1169-1174) suggests
the uniform distribution on the interval (7.5, 20) as a model for
depth (cm) of the bioturbation layer in sediment in a certain region.
a. What are the mean and variance of depth?
b. What is the cdf of depth?
7