Question

6. Some popular restaurants may have a very long line to get seated. Let random variable X denote your waiting time in hours). We suppose that X satisfies an exponetial distribution with parameter = 1. (a) What is the probability of being waiting for at least an hour? (b) Let's supposed you have waited for 1 hour. What is the conditional probability of waiting for 1 more hour given you have waited for 1 hour? (c) What is the expected waiting time given you have waited for 1 hour?

Answer 1

for exponential distribution:

F(x)=P(X<x)=1-e-λx

a)

P(being waiting for at least an hour) =P(X>1) =1-P(X<1) =1-(!-e^-1*1) =e^-1 =0.3679

b)P(waiiting more than 2 hours given waited for more than 1 hours)

P(X>2 |X>!)=P(X>2)/P(X>1 )=e^-1*2/e^-1*1 =e^-1 =0.3679

c)

expected waiting time given X>1 =1+E(X|X>1) = 1+$\int_{1}^{\infty }$ x*f(x|x>1) dx =1+$\int_{1}^{\infty }$ x*e^-(x-1) dx =1+(-xe^-(x-1)-e^-(x-1) )|^$\infty$₁

=1+1 =2 hours