Question

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​?

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 19.8, \sigma = 5.1$

a) About what percent of students scored over 32​?

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 19.8}{5.1}$

$Z = 2.39$

$Z = 2.39$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 19.8}{5.1}$

$Z = -0.55$

$Z = -0.55$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 19.8}{5.1}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

X = 17

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 19.8}{5.1}$

$Z = -0.55$

$Z = -0.55$ has a pvalue of 0.2912

0.9916 - 0.2912 = 0.7004

70.04% of students scored between 17 and 32.