Public transit deaths. Millions of suburban commuters use the public transit system (e.g., subway trains) as an alter native to the automobile. While generally perceived as a safe mode of transportation, the average number of deaths per week due to public transit accidents is 5 (Bureau of Transportation Statistics, 2015).

a. Construct arguments both for and against the use of the Poisson distribution to characterize the number of deaths per week due to public transit accidents.

b. For the remainder of this exercise, assume the Poisson distribution is an adequate approximation for x, the number of deaths per week due to public transit accidents. Find $\mathit{E}\left(x\right)$and the standard deviation of x.

c. Based strictly on your answers to part b, is it likely that more than 12 deaths occur next week? Explain.

d. Find$\mathit{p}\left(x>12\right)$. Is this probability consistent with your answer to part c? Explain.

Let x be the random variable that the number of death per week due to public transit accidents.

Requirements for the Poisson distribution

1. The occurrences must be uniformly distributed for the given interval,
2. The occurrences should be independent of each other.

Condition 1:

If the number of deaths follows the Poisson distribution, the probability of death that occurs per week must be same for every week.

Condition 2:

The occurrence of death in any week should be independent of any other week.

We can conclude that the 1^st assumption is not valid because the chance of occurrence of the death is not same for every week. This implies that there may be difference in the number of passengers for week to week.

Hence it is clear that the probability of death may also differ. Moreover, the seasons differ with the period which may results in decrease or increase of the accidents.

Given that the average number of deaths per week due to public transit accident is 5 which follows Poisson distribution with parameter$\lambda =5$

Hence the value of mean is

$$\begin{gathered} \mu =\lambda \\ =5 \end{gathered}$$

Therefore, the value of$E\left(x\right)$ is 5.

The standard deviation of x can be given by

$$\begin{gathered} \sigma =\sqrt{\lambda } \\ =\sqrt{5} \\ =2.2361 \end{gathered}$$

Hence the standard deviation of x is 2.2361

c.

from part (b) $\mu =5$and$\sigma =2.2361$

the Z-score for x=12 can be obtained as

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{12-5}{2.2361} \\ =\frac{7}{2.2361} \end{gathered}$$

That is

Hence, the value of z-score is 3.13.

The z-score that is 3.13 is more than 3 standard deviations away from the mean

which indicates that is very unlikely that more than 12 deaths occur for next week.

d.

We have to calculate the$\mathrm{P}\left(\mathrm{x}>12\right)$

By using the Poisson distribution with mean as 5

Value of$\mathrm{P}\left(\mathrm{x}\le 12\right)=0.997981$

Hence

$$\begin{gathered} \mathrm{P}\left(\mathrm{x}>12\right)=1-\mathrm{P}\left(\mathrm{x}\le 12\right) \\ =1-0.997981 \\ =0.0020 \end{gathered}$$

Hence the value of$\mathrm{P}\left(\mathrm{x}>12\right)$ is 0.0020

The probability of$\mathrm{P}\left(\mathrm{x}>12\right)$ is calculated as 0.0020, which is very small and consistent with the answer of part (c) by indicating it is very unlikely that more than 12 deaths occurs for next week.