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