Suppose the mean income of firms in the industry for a year is 90 million dollars with a standard deviation of 15 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 109 million dollars? Round your answer to four decimal places.

Answer:

0.8980 = 89.80% probability that a randomly selected firm will earn less than 109 million dollars

Step-by-step explanation:

Normal Probability Distribution:

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

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

Distributed normally, mean of 90 million, standard deviation of 15 million.

This means that [tex]\mu = 90, \sigma = 15[/tex]

What is the probability that a randomly selected firm will earn less than 109 million dollars?

This is the pvalue of Z when X = 109. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{109 - 90}{15}[/tex]

[tex]Z = 1.27[/tex]

[tex]Z = 1.27[/tex] has a pvalue of 0.8980

0.8980 = 89.80% probability that a randomly selected firm will earn less than 109 million dollars

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