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.

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 and standard deviation , the z-score of a measure X is given by:

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

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

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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step-by-step explanation:

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-12 -5(x + 4) = 3x

-12 - 5x - 20 = 3x

-12 - 20 - 5x = 3x

-32 - 5x = 3x

    + 5x   + 5x

-32 = 8x

8         8

- 4= x