The mathematics section of a standardized college entrance exam had a mean of 19.8 and an SD of 5.1 for a recent year. Assume these are well modeled by a Normal distribution. a) About what percent of students scored over 32? b) About what percent of students scored under 17? c) About what percent of students scored between 17 and 32?

a) 0.84% of students scored over 32

b) 29.12% of students scored under 17

c) 70.04% of students scored between 17 and 32.

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:

a) About what percent of students scored over 32?

This is 1 subtracted by the pvalue of Z when X = 32. So

has a pvalue of 0.9916

1 - 0.9916 = 0.0084

So 0.84% of students scored over 32

b) About what percent of students scored under 17?

This is the pvalue of Z when X = 17. So

has a pvalue of 0.2912

So 29.12% of students scored under 17

c) About what percent of students scored between 17 and 32?

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 12. So

X = 32

has a pvalue of 0.9916

X = 17

has a pvalue of 0.2912

0.9916 - 0.2912 = 0.7004

70.04% of students scored between 17 and 32.

this doesn't make sense lol

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step-by-step explanation: