Question

Entry to a certain University is determined by a national test. The scores on this test are known to be normally distributed with a mean of 725 and a standard deviation of 115. Harriet wants to be admitted to this university and she knows that she must score better than at least 75% of the students who took the test. Harriet takes the test and scores 874.5. 1. (2pts) What percentile did Harriet score in? 2. (2pt) Will Harriet be admitted? Wh

Answer 1

She scored in the 90th percentile.

90th percentile is better than 90% of the students who took the test, so yes, she will be admitted.

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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.

The scores on this test are known to be normally distributed with a mean of 725 and a standard deviation of 115.

This means that
`\mu = 725, \sigma = 115`

She must score better than at least 75% of the students who took the test.

So her z-score should have a pvalue of at least 0.75.

Harriet takes the test and scores 874.5. 1. (2pts) What percentile did Harriet score in?

We have to find the pvalue of Z when X = 874.5.

`Z = (X - \mu)/(\sigma)`

`Z = (874.5 - 725)/(115)`

`Z = 1.3`

`Z = 1.3` has a pvalue of 0.9032

She scored in the 90th percentile.

Will Harriet be admitted?

90th percentile is better than 90% of the students who took the test, so yes, she will be admitted.