235k views
3 votes
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?

User Emmanu
by
7.7k points

1 Answer

3 votes

Answer:

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.

User Nicol
by
7.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories