Question

Recall that the pdf for the combined waiting times for two buses from HW 4.1 Q3 was f(y) = {1/ 25 y ( 0 ≤ y < 5), 2/ 5 − 1/ 25 y (5 ≤ y ≤ 10) , 0 (y < 0 or y > 10 a)}. Compute the cdf of Y. Note that you will have to find a constant C that you add. For 0 ≤ y < 5, let C = 0, so that you only need to find C for 5 ≤ y ≤ 10. To find it, remember that F(∞) = 1 and that f(y) = 0 for y > 10, so that y > 10 contributes nothing more to the cdf. This means F(10) must equal 1. b. Obtain an expression for the (100p)th percentile. (Hint: consider separately 0 < p < .5 and .5 < p < 1.) c. Find µ˜, the median, using the 50th percentile. d. Compute µY and σ 2 Y . How do these compare with the expected waiting time and variance for a single bus when the time is uniformly distributed as follows? f(x) = ( 1 5 0 ≤ x ≤ 5 0 else

Answer 1

Solution:

a. When 0 y 5,

F (y) = = y^2/50

For 5 y 10,

F (y) =

=

= 1/2 +

= 2/5y - y^2/50 - 1

b. For 0 < p 0.5, p = F () = /50

= (50p)^1/2

For 0.5 < p   1, p = 2/5 - /50 - 1

= 10 - 52 (1- p)

c. E (Y) = 5 by straightforward integration and similarly V (Y) = 50/12 = 4.1667

The waiting time X for a single bus follows U (0, 5) and so

E (X) = 2.5 and V (X) = 25/12

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