Hw3-solution
3.6
Possible X values are1, 2, 3, 4, … (all positive integers)
Outcome: RL AL RAARL RRRRL AARRL
X: 2 2 5 5 5
3.10
a. T = total number of pumps in use at both stations. Possible values: 0, 1, 2, 3, 4, 5, 6, 7,
8, 9, 10
b. X: -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6
c. U: 0, 1, 2, 3, 4, 5, 6
d. Z: 0, 1, 2
3.12
a. In order for the flight to accommodate all the ticketed passengers who show up, no more
than 50 can show up. We need Y ≤ 50.
P(Y ≤ 50) = .05 + .10 + .12 + .14 + .25 + .17 = .83
b. Using the information in a. above, P(Y > 50) = 1 - P(Y ≤ 50) = 1 - .83 = .17
c. For you to get on the flight, at most 49 of the ticketed passengers must show up. P(Y ≤
49) = .05 + .10 + .12 + .14 + .25 = .66. For the 3rd person on the standby list, at most 47
of the ticketed passengers must show up. P(Y ≤ 47) = .05 + .10 + .12 = .27
3.14
a.
5
1
)(
y
yp = K[1 + 2 + 3 + 4 + 5] = 15K = 1 15
1
K
b. P(Y 3) = p(1) + p(2) + p(3) = 4.
15
6
c. P( 2 Y 4) = p(2) + p(3) + p(4) = 6.
15
9
d. 1
50
55
]2516941[
50
1
50
5
1
2
y
y; No
