Question

On average, 6 customers make purchases every 2 hours at a certain hardware. Assume that the number of customers making purchases follows a Poisson distribution. If a customer has just made a purchase, calculate (to 4 decimal places) the probability that the time until the next customer arrives to make a purchase is less than 5 minutes.

Answer 1

Solution:

If X, the number of purchase, follows Poisson distribution with parameter 6 . Then the probability mass function of X is

$P(X=x)=\frac{e^{-6} 6^{x}}{x !} \quad ; x=0,1,2 \ldots . .$

Then the time T for the occurrence of an event follows exponential distribution with parameter 6 . And the corresponding probability density function is

$f_{T}(t)=6 e^{-6 t} \quad ; \quad t>0$

And the corresponding cumulative distribution function is

$F_{T}(t)=P(T \leq t)=1-e^{-6 t}$

The Poisson parameter is given for hours. So $5 \mathrm{~min}=\frac{1}{24}(2 hours )$

so the required probability is

$F_{T}\left(\frac{1}{24}\right)=P\left(T \leq \frac{1}{24}\right)=1-e^{\frac{-6}{24}}=0.2212$