Question

Suppose the mean income of firms in the industry for a year is 90 million dollars with a standard deviation of 5 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 94 million dollars? Round your answer to four decimal places.

Answer 1

Answer:

0.7881 = 78.81% probability that a randomly selected firm will earn less than 94 million dollars

Step-by-step explanation:

Problems of normally distributed samples are 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:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 90, \sigma = 5$

If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 94 million dollars?

This is the pvalue of Z when X = 94. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{94 - 90}{5}$

$Z = 0.8$

$Z = 0.8$ has a pvalue of 0.7881

0.7881 = 78.81% probability that a randomly selected firm will earn less than 94 million dollars