Suppose the mean income of firms in the industry for a year is 90 million dollars with a standard deviation of 5 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 94 million dollars? Round your answer to four decimal places.

0.7881 = 78.81% probability that a randomly selected firm will earn less than 94 million dollars

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 the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 94 million dollars?

This is the pvalue of Z when X = 94. So

has a pvalue of 0.7881

0.7881 = 78.81% probability that a randomly selected firm will earn less than 94 million dollars

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.