Answer:
The standard deviation for annual precipitation in St. Louis is 6.48 inches.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 34[/tex]
If 17.75% of the time, the annual precipitation is more than 40 inches, what is the standard deviation for annual precipitation in St. Louis
This means that Z when X = 40 has a pvalue of 1-0.1775 = 0.8225. So when X = 40, Z = 0.925.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.925 = \frac{40 - 34}{\sigma}[/tex]
[tex]0.925\sigma = 6[/tex]
[tex]\sigma = \frac{6}{0.925}[/tex]
[tex]\sigma = 6.48[/tex]
The standard deviation for annual precipitation in St. Louis is 6.48 inches.
