Answer to Question #231668 in Statistics and Probability for tam

Question

The number of new smart phones that a phone shop sells follows a

Poisson distribution with a mean of 2 smart phones sold per day.

(i) Find the probability that the shop sells at least 4 new smart phones

during a period of 3 days. (4 marks)

(ii) Find the probability that the shop sells less than 65 new smart

phones in the month of September. (4 marks)

Answer 1

Poisson distribution has the formula:

$P(X=k)=\frac{\lambda^k\cdot e^{-\lambda}}{k!}$ ,

where E(X)=$\lambda$ =2.

We have $P(X\ge4)=1-P(X=0)-P(X=1)-P(X=2)-P(X=3)=1-e^{-\lambda}\cdot (\lambda ^0/0!+\lambda ^1/1!+\lambda ^2/2!+\lambda ^3/3!)=\\ =1-\frac{2^0/0!+2^1/1!+2^2/2!+2^3/3!}{e^2}=1-\frac{1+2+2+4/3}{e^2}=\\ =1-\frac{19}{3\cdot e^2}=0.143;$

(ii) We use sum Y of 30 random variable with Poisson distribution which has Poisson distribution also but with mean $\lambda$ =$30\cdot2=60$ and approximate it by normal distribution $N(60,60)$ . Then $P(0\le Y\le65)=\Phi(\frac{65-60}{\sqrt{60}})+\Phi(60/\sqrt{60})=\\ \Phi(0.65)+\Phi(7.75)=0.2422+0.5=0.7422$ where

$\Phi(x)$ - integral function of Laplace.