Answer to Question #110106 in Statistics and Probability for Emmanuel Daniels

Question

At a checkout counter customers arrive at an average of 2 per minute. Find the
probability that
(a) at most 4 will arrive at any given minute.
(b) at least 3 will arrive during an interval of 2 minutes
(c) 5 will arrive in an interval of 3 minutes.

Answer 1

(a) Poisson distribution with $\mu=2.$

$P(X\le4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=$

$=e^{-\mu}(\frac{\mu^0}{0!}+\frac{\mu^1}{1!}+\frac{\mu^2}{2!}+\frac{\mu^3}{3!}+\frac{\mu^4}{4!})=0.1429.$

(b) Poisson distribution with $\mu=2*2=4.$

$P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2)=$

$1-e^{-4}(\frac{\mu^0}{0!}+\frac{\mu^1}{1!}+\frac{\mu^2}{2!})=0.7619.$

(c) Poisson distribution with $\mu=2*3=6.$

$P(X=5)=e^{-mu}\frac{\mu^5}{5!}=0.1606.$