Question

On average, 3 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 7 minutes:

Answer 1

$\begin{aligned} \lambda &=\frac{3}{2 \times 60} \\ &=3 / 120 \\ \lambda &=0.025 \end{aligned}$

By using Poisson distribution $P(X=x)=\frac{e^{-\lambda} \cdot \lambda^{x}}{x!}$ ,

$\begin{aligned} P(X<7) &=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6) \\ &=0.95731+0.02438+0.00330+0.0000+00000+0.000+0.0000 \\ P(X<7) &=1.0000 \end{aligned}$