Question

The number of customers arriving per hour at a certain automobile service facility is assumed to follow a Poisson distribution with mean λ=5.

a) Compute the probability that more than 11 customers will arrive in a 2​-hour period.

b) What is the mean number of arrivals during a 2​-hour ​period?

Answer 1

Here, λ = 10 and x = 11
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X <= 11).

P(X <= 10) = (10^0 * e^-10/0!) + (10^1 * e^-10/1!) + (10^2 * e^-10/2!) + (10^3 * e^-10/3!) + (10^4 * e^-10/4!) + (10^5 * e^-10/5!) + (10^6 * e^-10/6!) + (10^7 * e^-10/7!) + (10^8 * e^-10/8!) + (10^9 * e^-10/9!) + (10^10 * e^-10/10!) + (10^11 * e^-10/11!)
P(X <= 10) = 0 + 0.0005 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 + 0.1137
P(X <= 10) = 0.6968

P(x > 11) = 1 - P(x <= 11)
= 1- 0.6968
= 0.3032

b)

the mean number of arrivals during a 2​-hour​period = 5 * 2 = 10