Using the normal distribution, it is found that there is a 0.0548 = 5.48% probability that, in a given year, the rainfall is over 40 inches.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem:
The probability that, in a given year, the rainfall is over 40 inches is 1 subtracted by the p-value of Z when X = 40, so:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{40 - 32}{5}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a p-value of 0.9452.
1 - 0.9452 = 0.0548.
0.0548 = 5.48% probability that, in a given year, the rainfall is over 40 inches.
A similar problem is given at https://brainly.com/question/24663213