Answer to Question #235041 in Statistics and Probability for hype

Question

Phone calls enter the ”support desk” of an electricity supplying company on the average two every 3 minutes. If one assumes an approximate Poisson process: (i) What is the probability of no calls in 3 minutes? [2 marks] (ii) What is the probability of utmost 6 calls in a 9 minute period? [2 marks]

Answer 1

$p(x)=\frac{\lambda^xe^{-\lambda}}{x!}, x= 0,1,2,3,...$

i. Probability of no call in 3 minutes

$\lambda=2$

$p(0)=\frac{2^0e^{-2}}{0!}$

$=e^{-2}=0.13534$

ii. Probability of utmost 6 calls in 9 minutes

$\lambda =2\times\frac{9}{3}=6$

$p(\le6)=\sum_0^6\frac{6^xe^{-6}}{x!}$

$=\frac{6^0e^{-6}}{0!}+\frac{6^1e^{-6}}{1!}+\frac{6^2e^{-6}}{2!}+...+\frac{6^6e^{-6}}{6!}$

$=0.00248+0.014873+0.044618+...+0.16062$

$=0.6063$