Question

Customers arrive at a checkout counter in a department store according to a Poison distribution of an average of seven per hour.During a given hour,what are the probabilities that

(a).no more than three customers arrive?

(b).at least two customers arrive?

(c).exactly four customers arrive?

Answer 1

Given, $\lambda=7$

Let X denote the number of customers arrive.

(i) $P(X\le 3)=P(X=1)+P(X=2)+P(X=3)+P(X=4)$

$=\dfrac{e^{-\lambda}\lambda^0}{0!}+\dfrac{e^{-\lambda}\lambda^1}{1!}+\dfrac{e^{-\lambda}\lambda^2}{2!}+\dfrac{e^{-\lambda}\lambda^3}{3!}$

$=\dfrac{e^{-7}7^0}{0!}+\dfrac{e^{-7}7^1}{1!}+\dfrac{e^{-7}7^2}{2!}+\dfrac{e^{-7}7^3}{3!}$

$=0.000911+0.00638+0.0223+0.05212=0.08172$

(ii) $P(X\ge 2)=1-P(X<2)$

$=1-(P(X=0)+P(X=1))$

$=1-(\dfrac{e^{-\lambda}.\lambda^0}{0!}+\dfrac{e^{-\lambda}\lambda^1}{1!})$

$=1-(\dfrac{e^{-7}7^0}{0!}+\dfrac{e^{-7}7^1}{1!})$

$=1-(0.000911+0.0638)=0.9927$

(iii) $P(X=4)=\dfrac{e^{-\lambda}\lambda^4}{4!}$

$=\dfrac{e^{-7}7^4}{24}$

$=0.09122$