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

answer: the common difference is -0.10

answer: 7 x 2/5 = 2.8 hours, is your answer. your welcome! : 3

step-by-step explanation: first work out how long it took to hose 1/7 of the lawn which is 2/5 hours. times that by 7 to get the whole lawn.

sorry i don't know the other answer. i didn't learn what a proportional relationship is.