Based on an analysis of sample data an article proposed the

Based on an analysis of sample data, an article proposed the pdf

f(x) = 0.45e^{0.45(x 1)}

when

x 1

as a model for the distribution of

X = time (sec)

spent at the median line. (Round your answers to three decimal places.)

(a) What is the probability that waiting time is at most 6 sec? More than 6 sec?

(b) What is the probability that waiting time is between 8 and 9 sec?

Solution

For the exponential distribution

f(x) = A exp [-A(x - b)],

the cumulative distribution function is

F(x) = 1 - exp [-A(x-b)]

Thus, here,

F(x) = 1 - exp [-0.45(x-1)]

Thus,

a)

P(x<=6) = F(6) = 1 - exp [-0.45(6 - 1)] = 0.894600775 [answer]

P(x>6) = 1 - F(6) = 1 - {1 - exp [-0.45(6 - 1)]} = 0.105399225 [answer]

******************

b)

P(8<x<9) = F(9) - F(8) = {1 - exp [-0.45(9 - 1)]} - {1 - exp [-0.45(8 - 1)]} = 0.015528404 [answer]