Question

A small business owner determines that her revenue during the next year should be approximately normally distributed with a mean of $425,000 and a standard deviation of $130,000. What is the probability that her revenue will exceed $600,000?

Answer 1

8.85% probability that her revenue will exceed $600,000.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that

`\mu = 425000, \sigma = 130000`

What is the probability that her revenue will exceed $600,000?

This is 1 subtracted by the pvalue of Z when X = 600000.

So

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

`Z = (600000 - 425000)/(130000)`

`Z = 1.35`

`Z = 1.35` has a pvalue of 0.9115.

So there is a 1-0.9115 = 0.0885 = 8.85% probability that her revenue will exceed $600,000.