Question

Currently, an average of 7 residents of the village of Westport (population 760) die each year (based on data from the U.S. National Center for Health Statistics).

a. Find the mean number of deaths per day.

b. Find the probability that on a given day, there are no deaths.

c. Find the probability that on a given day, there is more than one death.

d. Based on the preceding results, should Westport have a contingency plan to handle more than one death per day? Why or why not?

Short answer

a. The mean number of deaths is 0.0192.

b.The probability that on a given day, there is no death is 0.98.

c. The probability that on a given day, there is more than one death is 0.000182.

d. No, Westport does not need to have a contingency plan to handle more than one death per day.

Step-by-step solution

Given information

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

Step 2: Compute the mean number of deaths per day

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

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

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

Compute the probability for no death 

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

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

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

Compute the probability that there is more than one death

c.

The probability of more than one death is given as

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

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

Step 5: Comment if Westport should have a contingency plan to handle more than one death per day

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.