Question

Service calls come to a maintenance center according to a Poisson process, and on

average, 2.7 calls are received per minute. Find the probability that

a) no more than 4 call in any minute;

b) fewer than 2 calls come in any minute;

c) more than 10 calls come in a 5-minute period.

Answer 1

We have a Poisson distribution,

$\lambda=2.7; P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!}=\cfrac{(2.7t)^k\cdot e^{-2.7t}}{k!}.$

a)

$P_1(X\leq4)=\\ =P_1(X=0)+P_1(X=1)+P_1(X=2)+\\ +P_1(X=3)+P_1(X=4)=\\ =\cfrac{2.7^0\cdot e^{-2.7}}{0!}+\cfrac{2.7^1\cdot e^{-2.7}}{1!}+\\ +\cfrac{2.7^2\cdot e^{-2.7}}{2!}+\cfrac{2.7^3\cdot e^{-2.7}}{3!}+\\ +\cfrac{2.7^4\cdot e^{-2.7}}{4!}=\\ =0.0672+0.1815+0.2450+0.2205+0.1488=\\=0.8629.$

b)

$P_1(X<2)=\\ =P_1(X=0)+P_1(X=1)=\\ =0.0672+0.1815=0.2487.$

c)

$\lambda t=2.7\cdot5=13.5;\\ P_5(X>10)=1-P_5(X\leq10)=\\ =1-(P_5(X=0)+P_5(X=1)+...+P_5(X=10))=\\ =1-\begin{pmatrix} \cfrac{13.5^0}{0!}+\cfrac{13.5^1}{1!}+...\cfrac{13.5^{10}}{10!} \end{pmatrix}\cdot e^{-13.5}=\\ =1-(1+13.5+91.13+410.06+1383.96+3736.70+\\ +8407.56+16214.59+27362.11+41043.13+\\ +55408.28)\cdot e^{-13.5}=\\ =1-0.2112=0.7888.$