a.The probability of 7 hurricanes in a year is equal to 0.140.

b. Thus, the expected number of years to have 7 hurricanes in a 55-year period is equal to 7.7 years.

c. The expected number of years that have 7 hurricanes is approximately equal to the actual number of years that have 7 hurricanes in a 55-year period.Since the expected and the actual values are approximately equal, the Poisson distribution works well here.

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

a.

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

The probability of 7 hurricanes in a year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 7 \right) = \frac{{{{\left( {6.1} \right)}^7}{{\left( {2.71828} \right)}^{ - 6.1}}}}{{7!}}\\ = 0.139856\\ \approx 0.140\end{aligned}\]

Therefore, the probability of 7 hurricanes in a year is equal to 0.140.

b.

The expected number of years to have 7 hurricanes in a 55-year period is computed below:

\(\begin{aligned}{c}55 \times P\left( 7 \right) = 55 \times 0.140\\ \approx 7.7\end{aligned}\)

Thus, the expected number of years to have 7 hurricanes in a 55-year period is equal to 7.7 years.

c.

It is given that the actual number of years that had 7 hurricanes in the recent 55-year period is equal to 7.

The expected number of years that have 7 hurricanes in a 55-year period is equal to 7.7.

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

Since the expected and the actual values are approximately equal, the Poisson distribution works well here.