Question

The number of 911 calls in Washington DC, has a Poisson distribution with a mean of 8 calls a day. Find

a.The probability of eight 911 calls in a day

b.The probability of at most two 911 calls in a day

c.The probability of some 911 calls in a day

Answer 1

$We \ have, \ X\sim Poisson(8)\\ P(X=x)=\frac{e^{-8}8^x}{x!},x=0,1,2...\\ a. P(8 \ calls \ in \ a \ day)=P(X=8)=\frac{e^{-8}8^8}{ 8!}=0.139587\\ b. P(at \ most \ 2\ calls \ in \ a \ day)=P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)\\ =\frac{ e^{-8}8^0}{0!}+\frac{e^{-8}8^1}{ 1!}+\frac{e^{-8}8^2}{ 2!}=0.000335463+0.0026837+0.0107348=0.013754\\ c) P(some \ calls \ in \ a \ day)=1-P(no \ calls \ in \ a\ day)=1-P(X=0)=1-\frac{ e^{-8}8^0}{0!}=0.999665$