Question

Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is the probability that a student scores between 350 and 550? b. If the upper 3% scholarship, what score must a student receive to get a scholarship? c. Find the 60th percentile of the test scores. d. Find the middle 30% of the test scores.

Answer 1

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

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 = (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 = 450, \sigma = 100`

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

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

`Z = (550 - 450)/(100)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

X = 350

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

`Z = (350 - 450)/(100)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

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

`1.88 = (X - 450)/(100)`

`X - 450 = 1.88*100`

`X = 638`

A score of 638(or higher).

c. Find the 60th percentile of the test scores.

X when Z has a pvalue of 0.60. So it is X when Z = 0.253

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

`0.253 = (X - 450)/(100)`

`X - 450 = 0.253*100`

`X = 475.3`

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

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

`-0.385 = (X - 450)/(100)`

`X - 450 = -0.385*100`

`X = 411.5`

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

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

`0.385 = (X - 450)/(100)`

`X - 450 = 0.385*100`

`X = 488.5`

The middle 30% of the test scores is between 411.5 and 488.5.