Question

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

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
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 80, \sigma = 7`

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

`Z = (X - \mu)/(\sigma)`

`Z = (85 - 80)/(7)`

`Z = 0.71`

`Z = 0.71` 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