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 at most 3 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(atmost 3 accident) = P(X ≤ 3)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= `[("e"^-4(4)^0)/(0!) + "e"^-4 (4)^1/(1!)  + ("e"^-4(4)^2)/(2!) + ("e"^-4(4)^3)/(3!)]`

= `"e"^-4 [4^0/(0!)  + 4^1/(1!) + 4^2/(2!) + 4^3/(3!)]`

= `0.0183 [1 + 4 + 16/2 + 64/6]`

= `0.0183 [1 + 4 8 + 32/3]`

= `0.0183[1 + 2/3]`

= `0.0183[(9 + 32)/3]`

= `0.0061[71]`

= 0.4331

Out of 100 days there will be at most 3 acccident = n × P(X ≤ 3)

= 100 × 0.4331

= 43.31

= 43 days  ........(approximately)