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 -------.

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Answer:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 72, \sigma = 12[/tex]

Percentage of students who get at least 78:

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{78 - 72}{12}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] 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

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