Question

The annual rainfall in Frostbite Falls is normally distributed with mean

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 32 inches, so $\mu = 32$.
• Standard deviation of 5 inches, so $\sigma = 5$.

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:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{40 - 32}{5}$

$Z = 1.6$

$Z = 1.6$ 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 brainly.com/question/24663213