The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 90 inches comma and a standard deviation of 12 inches. what is the probability that the mean annual snowfall during 36 randomly picked years will exceed 92.8 inches question mark round your answer to four decimal places.

Step-by-step

standard deviation (sigma) = 12 inches

mean (mu) = 90 inches.

For 36 years, the standard error of the mean is

e = sigma / sqrt(n) = 12" / sqrt(36) = 2".

Deviation from the mean

= 92.8 -90 = 2.8

Since the standard error of the mean follows the normal distribution, we have

Z=(92.8-90)/2 = 1.4

Looking up the normal distribution tables, or check R, we find

P(Z>1.4) = 1-P(Z<1.4) = 1-0.9192433 = 0.08075666

P(snowfall>92.8) = 0.0808 (to four decimal places)

idk

step-by-step explanation: