Question

The distribution of the number of road accidents per day in a city is poisson with mean 4. Find the number of days out of 100 days when there will be atleast 2 accidents 

Answer 1

In a possion distribution

Mean λ = 4

n = 100

x follows possion distribution with

P(x) = `("e"^(-lambda) lamda^x)/(x!)`

= `("e"^-4 (4)^x)/(x!)`

P(atleast 2 accidents)

= P(X ≥ 2)

= P(X = 2) + P(X = 3) + P(X = 4) + …………

= 1 – P(X < 2)

= 1 – [P(X = 0) + P(X = 1)]

= `1 - [("e"^-4(4)^0)/(0!) + ("e"^-4(4)^1)/(1!)]`

= 1 – e^-4[l + 4]

= 1 – 0.0183(5)

= 1 – 0.0915

= 0.9085

= Out of 100 days there will be atleast 2 accidents

= n × P(X ≥ 2)

= 100 × 0.9085

= 90.85

= 91 days .......(approximately)