Respuesta :
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]
- 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 [tex]\mu = 32[/tex].
- Standard deviation of 5 inches, so [tex]\sigma = 5[/tex].
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
