Answer to Question #284048 in Statistics and Probability for Nery

Question

Let the number of car accidents be a Poisson R.V. If the rate of car accidents in a road is 3 accidents in a day , compute () the probability of having no accidents in the afternoon interval (12 pm, 18pm). ( ii) the average and standard deviation of accidents in a week.

Answer 1

Solution:

Let x= Number of car accidents.

rate =3 accidents in a day.

(i) There are 24 hours in a day.

Hence rate of accidents per hours $=\frac{3}{24}$

$\begin{aligned} \Rightarrow \quad & x \sim \operatorname{Poisson}(\lambda) \\ & x \sim \operatorname{Poisson}\left(\frac{3}{24}\right) . \\ \text { Time interval; } t=t_{2}-t_{1} \\ & t=181 \mathrm{H}-12 \mathrm{PM}=6 \text { hours} \\ \Rightarrow \quad & x \sim \operatorname{Poisson}(\lambda t) \\ & x \sim \operatorname{Poisson}\left(\frac{3}{24} \times 6\right) \sim P\left(\frac{3}{4}\right) \\ & \text { hence } \lambda t=\frac{3}{4} . \end{aligned}$

P[no accident in the interval 12 PM to 18 PM] =

$\begin{aligned} &P[x=x]=\frac{e^{-\lambda t}(\lambda t)^{x}}{x !} \\ &P[x=0]=\frac{e^{-3 / 4}\left(\frac{3}{4}\right)^{0}}{0 !}=e^{-\frac{3}{4}}=0.47 \end{aligned}$

(ii) There are 7 days in a week

$\Rightarrow \lambda=3,t=7 \\\Rightarrow \lambda t=21 \\ X \sim Poisson(\lambda t)$

Average $=E(X)=\lambda t=21$ accidents per week.

Standard deviation $=\sqrt{\operatorname{var}(x)}=\sqrt{\lambda t}=\sqrt{21}$

$\Rightarrow \sigma=4.58$