Question

The Arc Electronic Company had an income of 80 million dollars last year. Suppose the mean income offirms in the same industry as Arc for a year is 100 million

dollars with a standard deviation of 15 million dollars. If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will
carn more than Arc did last year? Round your answer to four decimal places

Answer 1

23.89% probability that a randomly selected firm will earn more than Arc did last year

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation, the 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 value, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will earn more than Arc did last year?

This is 1 subtracted by the value of Z when X = 85. So

has a value of 0.7611

1 - 0.7611 = 0.2389

23.89% probability that a randomly selected firm will earn more than Arc did last year