Question

The amount of snowfall falling in a certain mountain range is normally distributed with a average of 170 inches, and a standard deviation of 20 inches. What is the probability a randomly selected year will have an average snofall above 200 inches

Answer 1

0.0668 = 6.68% probability a randomly selected year will have an average snowfall above 200 inches.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a average of 170 inches, and a standard deviation of 20 inches.

This means that
`\mu = 170, \sigma = 20`

What is the probability a randomly selected year will have an average snowfall above 200 inches?

This is 1 subtracted by the p-value of Z when X = 200. So

`Z = (X - \mu)/(\sigma)`

`Z = (200 - 170)/(20)`

`Z = 1.5`

`Z = 1.5` has a p-value of 0.9332.

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability a randomly selected year will have an average snowfall above 200 inches.