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) = ${\displaystyle \frac{{\text{e}}^{-\lambda }{\lambda }^x}{x!}}$

= ${\displaystyle \frac{{\text{e}}^{-4}{\left(4\right)}^x}{x!}}$

P(atmost 3 accident) = P(X ≤ 3)

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

= ${\displaystyle [\frac{{\text{e}}^{-4}{\left(4\right)}^0}{0!}+{\text{e}}^{-4}\frac{{\left(4\right)}^1}{1!} +\frac{{\text{e}}^{-4}{\left(4\right)}^2}{2!}+\frac{{\text{e}}^{-4}{\left(4\right)}^3}{3!}]}$

= ${\displaystyle {\text{e}}^{-4}[\frac{4^0}{0!} +\frac{4^1}{1!}+\frac{4^2}{2!}+\frac{4^3}{3!}]}$

= ${\displaystyle 0.0183[1+4+\frac{16}{2}+\frac{64}{6}]}$

= ${\displaystyle 0.0183[1+48+\frac{32}{3}]}$

= ${\displaystyle 0.0183[1+\frac{2}{3}]}$

= ${\displaystyle 0.0183\left[\frac{9+32}{3}\right]}$

= ${\displaystyle 0.0061\left[71\right]}$

= 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)