Hurricanes. In Exercises 5–8, assume that the Poisson distribution applies; assume that the mean number of Atlantic hurricanes in the United States is 6.1 per year, as in Example 1; and proceed to find the indicated probability.

Hurricanes

a. Find the probability that in a year, there will be no hurricanes.

b. In a 55-year period, how many years are expected to have no hurricanes?

c. How does the result from part (b) compare to the recent period of 55 years in which there were no years without any hurricanes? Does the Poisson distribution work well here?

The mean number of Atlantic hurricanes in the United States is equal to 6.1 per year.

a.

Let X be the number of Atlantic hurricanes in one year. Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 6.1\).

The probability of no hurricanein a year is computed below:

\(\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 0 \right) = \frac{{{{\left( {6.1} \right)}^0}{{\left( {2.71828} \right)}^{ - 6.1}}}}{{0!}}\\ = 0.00224\end{aligned}\)

Therefore, the probability of no hurricane in a year is equal to 0.00224.

b.

The expected number of years to have no hurricane in a 55-year period is computed below:

\(\begin{aligned}{c}55 \times P\left( 0 \right) = 55 \times 0.00224\\ = 0.1232\\ \approx 0.12\end{aligned}\)

Thus, the expected number of years to have no hurricane in a 55-year period is equal to 0.12 years.

c.

It is given that the actual number of years that had no hurricanes in the recent 55-year period is equal to 0.

The expected number of years that have no hurricane in a 55-year period is equal to 0.12.

Thus, the expected number of years is approximately equal to the actual number of years.

Since the expected and the actual values are approximately equal, the number of hurricanes is well-modeled by the Poisson distribution.