The data for the MBA Statistics final exam scores is a random variable from a normal distribution with mean 72 and standard deviation 12. If 90 students take the course, the average number of students who get at least 78 is .

If 90 students take the course, the average number of students who get at least 78 is 27.765

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:

Percentage of students who get at least 78:

1 subtracted by the pvalue of Z when X = 78.

has a pvalue of 0.6915

1 - 0.6915 = 0.3085

If 90 students take the course, the average number of students who get at least 78 is

30.85% of 90. So

0.3085*90 = 27.765.

If 90 students take the course, the average number of students who get at least 78 is 27.765

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2