Question

A local drugstore owner knows that, on average, 100 people enter his store each hour.

a) Find the probability that in a given 3-minute period nobody enters the store.

b) Find the probability that in a given 3-minute period more than 5 people enter the

store.

Answer 1

We have a Poisson distribution,

$\lambda=100; P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!};\\ 3\ \rm{mins=\cfrac{3}{60} \ hour=0.05\ hour;}$

$\rm For\ a\ 3-minute\ period \ \lambda t=100\cdot0.05=5.$

a)

$P_{0.05}(X=0)=\cfrac{5^0\cdot e^{-5}}{0!}=0.0067.$

b)

$P_{0.05}(X>5)=1-P_{0.05}(X\leq5)=\\ =1-(P_{0.05}(X=0)+P_{0.05}(X=1)+\\ +P_{0.05}(X=2)+P_{0.05}(X=3)+\\ +P_{0.05}(X=4)+P_{0.05}(X=5))=\\ =1-\\ -\begin{pmatrix} \cfrac{5^0}{0!}+\cfrac{5^1}{1!}+\cfrac{5^2}{2!}+\cfrac{5^3}{3!}+\cfrac{5^4}{4!}+\cfrac{5^5}{5!}\end{pmatrix}\cdot e^{-5}=\\ =1-\\ -(1+5+12.5+20.83+26.04+26.04)\cdot e^{-5}=\\ =1-0.6160=0.3840.$