91Ó°ÊÓ

On average, 7 residents of the village of Westport (population 760) die each year.

a.

LetX be the number of residents of the village of Westport that die each day.

The mean value for X, that is, the number of deaths each day, is given as

$$\begin{gathered} \mathrm{μ}=\frac{\text{No}\text{of}\text{deaths}\text{per}\text{year}}{\text{Total}\text{number}\text{of}\text{days}\text{in}\text{a}\text{year}} \\ =\frac{7}{365} \\ =0.0192 \end{gathered}$$

Therefore, the mean number of deaths is equal to 0.0192.

b.

As the occurrences are random, independent, and uniformly distributed on the interval of 365 days of the year, the variable X would follow the Poisson distribution such that the mean is 0.0192.

The probability of no death is given as

$$\begin{gathered} P\left(x\right)=\frac{{\mathrm{μ}}^x{e}^{-\mathrm{μ}}}{x!} \\ P\left(0\right)=\frac{{\left(0.0192\right)}^0{\left(2.71828\right)}^{-0.0192}}{0!} \\ =0.981 \end{gathered}$$

Therefore, the probability that on a given day, there is no death is 0.981.

c.

The probability of more than one death is given as

$$\begin{gathered} P\left(x>1\right)=1-P\left(x=0\right)-P\left(x=1\right) \\ =1-\frac{{\left(0.0192\right)}^0{\left(2.71828\right)}^{-0.0192}}{0!}-\frac{{\left(0.0192\right)}^1{\left(2.71828\right)}^{-0.0192}}{1!} \\ =1-0.981-0.019 \\ ≈0.000182 \end{gathered}$$

Therefore, the probability that on a given day, there is more than one death is 0.000182.

d.

From the above results, the probability of more than one death is very small, that is, lesser than 0.05, which implies that the event is rare.

Therefore, Westport does not need to have a contingency plan to handle more than one death per day.