Question

Intelligence quotas on two different tests are normally distributed. Test A has a mean of 100 and a standard deviation of 13. Test B has a mean of 100 and a standard deviation of 18. Use​z-scores to determine which person has the higher​ IQ: an individual who scores 123 on Test A or an individual who scores 121 on Test B. Which individual has the higher​ IQ?

Answer 1

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.

Explanation:

Problems of normally distributed samples can be 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:

The individual with the higher IQ is the one with the higher z-score. So

Test A has a mean of 100 and a standard deviation of 13. An individual who scores 123 on Test A.

So
`\mu = 100, \sigma = 13, X = 123`

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

`Z = (123 - 100)/(13)`

`Z = 1.77`

Test B has a mean of 100 and a standard deviation of 18. An individual who scores 121 on Test B.

So
`\mu = 100, \sigma = 18, X = 121`

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

`Z = (121 - 100)/(18)`

`Z = 1.17`

The individual who scores 123 on test A has a higher zscore, so he has the higher IQ.