Question

(EXPONENTIAL DISTRIBUTION) Customers arrive at the claims counter at the rate of 20 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes? Hint: Compute P(X<5) 1-e after compute ] (3 pts.)

Answer 1

Here , the number of customer follows a Poisson( $\lambda t=20$ ) distribution.

where t = 60 min

So, $\lambda =20/60 = 1/3$

Then X~ Exp( $\lambda = 1/3$ )

So, P[X<5] = $\int_{0}^{5} \frac{1}{3}e^{-3x} dx$

$= \frac{1}{3} [-\frac {e^{-3x}}{3}]_{0}^{5}$

$= \frac{1}{9}[1-e^{-15}]$

= 0.111

If you have doubts, or queries regarding relation of exponential and Poisson rv s ask me below. Thanks