The Arc Electronic Company had an income of 80 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 85 million dollars with a standard deviation of 7 million dollars. 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

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

Step-by-step explanation:

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

In a set with mean and standard deviation , the zscore 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 pvalue, 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 pvalue of Z when X = 85. So

has a pvalue of 0.7611

1 - 0.7611 = 0.2389

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

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right