Question

Suppose the distribution of IQ scores of adults is normal with a mean of 100 and a standard deviation of 15. Find IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent. Round your answer to the nearest integer. The IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent is [IQValue].

Answer 1

The IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent is 107.

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.

In this question, we have that:

`\mu = 100, \sigma = 15`

Find IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent.

This is X when Z has a pvalue of 0.68. So X when Z = 0.47.

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

`0.47 = (X - 100)/(15)`

`X - 100 = 0.47*15`

`X = 107.05`

So, rounding to the nearest integer:

The IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent is 107.