Suppose the mean income of firms in the industry for a year is 75 million dollars with a standard deviation of 17 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 110 million dollars? Round your answer to four decimal places.

Question

Kadyn Kirby

Business

29 May, 00:37

Suppose the mean income of firms in the industry for a year is 75 million dollars with a standard deviation of 17 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 110 million dollars? Round your answer to four decimal places.

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Answers (1)

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Matthias Humphrey · Answer 1

Answer and Explanation:

Given:

μ = 75 million

SD = 17 million

Probability (x) raw data = 110 million

Computation:

= Probability (x) < 110 million

= Probability [ (x-μ) / SD] < [ (110 - 75) / 17]

[ (x-μ) / SD] = Z

= Probability [z] < [ (35) / 17]

= Probability [z] < [2.05882353]

Using z calculator:

P-value from Z-Table:

Z score = 0.98024

Therefore, probability is 0.98024