Based on data from a college, scores on a certain test are normally distributed with a mean of 1530 and a standard deviation of 322.

find the percentage of scores greater than 2317. (round to two decimal places as needed.)
find the percentage of scores less than 1190. % (round to two decimal places as needed.)
find the percentage of scores between 1351 and 1673.

0.73% of the scores are greater than 2317.

14.46% of the scores are less than 1190.

38.23% of the scores are between 1351 and 1673.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

Find the percentage of scores greater than 2317.

This is 1 subtracted by the pvalue of Z when X = 2317. So:

has a pvalue of 0.9927.

So 1-0.9927 = 0.0073 = 0.73% of the scores are greater than 2317.

Find the percentage of scores less than 1190.

This is the pvalue of Z when X = 1190. So:

has a pvalue of 0.1446.

So 14.46% of the scores are less than 1190.

Find the percentage of scores between 1351 and 1673.

This is the pvalue of Z when X = 1673 subtracted by the pvalue of Z when X = 1351. So

X = 1673

has a pvalue of 0.67

X = 1351

has a pvalue of 0.2877

So 0.67-0.2877 = 0.3823 = 38.23% of the scores are between 1351 and 1673.