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

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 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.

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

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.

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

x=12 units bd= 10 units

step-by-step explanation:

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation: