Answer to Question #323821 in Statistics and Probability for Bless

Question

 Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. During a given hour, what are the probabilities that

(a) no more than three customers arrive?

(b) at least two customers arrive?

(c) exactly four customers arrive? 

Answer 1

Let X= the number of costumers arrive at a checkout counter in a department store:X∼Po(λ).

$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$

$\lambda=7$

a. $P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=\frac{e^{-7}(7)^0}{0!}+\frac{e^{-7}(7)^1}{1!}+\frac{e^{-7}(7)^2}{2!}+\frac{e^{-7}(7)^3}{3!}=0.081765$

b. $P(X \ge 2)=1-(P(X=0)+P(X=1))=1-(\frac{e^{-7}(7)^0}{0!}+\frac{e^{-7}(7)^1}{1!})=0.992705$

c. $P(X=4)=\frac{e^{-7}(7)^4}{4!}=0.091226$