Question

The Arc Electronic Company had an income of 7676 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 9090 million dollars with a standard deviation of 77 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

Answer:

0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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:

$Z = \frac{X - \mu}{\sigma}$

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.

Suppose the mean income of firms in the same industry as Arc for a year is 90 million dollars with a standard deviation of 7 million dollars.

This means that $\mu = 90, \sigma = 7$

What is the probability that a randomly selected firm will earn more than Arc did last year?

Arc earned 76 million, so this is 1 subtracted by the pvalue of Z when X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 90}{7}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

1 - 0.0228 = 0.9772

0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year