Question 1205908
Let X denotes the number of persons arriving at the counter.

a Poisson distribution with parameter: {{{lambda=2}}} per {{{5}}} minutes

{{{P(x, lambda)=(e^(-lambda)*lambda^x)/x!}}}

{{{P(x,2)=(e^-2*2^x)/x!}}}


the probability function is

{{{f(x)= (e^-2*2^x)/x!}}}

The probability that {{{none}}} arrive in a five-minute period is 

{{{f(0)= (e^-2*2^0)/0!=e^(-2)}}}

{{{P(X=0)=e^(-2)}}}

{{{P(X=0)=1/e^2}}}

{{{P(X=0)=0.1353}}}



The probability that {{{more }}}{{{than }}}{{{four}}} customers arrive in a {{{5}}} minute period is

{{{P(X>4)=1-P(X<=4)}}}

{{{P(X>4)=1-(f(0)+f(1)+f(2)+f(3)+f(4))}}}

{{{f(0)= (e^-2*2^0)/0!=e^-2}}}
{{{f(1)= (e^-2*2^1)/2!=2e^-2/2=e^-2}}}
{{{f(2)= (e^-2*2^2)/2!=4e^-2/2=2e^-2}}}
{{{f(3)= (e^-2*2^3)/3!=8e^-2/6=4e^-2/3}}}
{{{f(4)= (e^-2*2^4)/4!=16e^-2/24=2e^-2/3}}}


{{{P(X>4)=1-(e^-2+e^-2+2e^-2+4e^-2/3+2e^-2/3)}}}

{{{P(X>4)=1-6e^-2}}}

{{{P(X>4)=0.1879883}}}

{{{P(X>4)=0.1880}}}