Question

Assume that adults have IQ scores that are normally distributed with a mean of 102.6 and a standard deviation of 23.6. Find the probability that a randomly selected adult has an IQ greater than 134.1.

Answer 1

9.18% probability that a randomly selected adult has an IQ greater than 134.1

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 = 102.6, \sigma = 23.6`

Find the probability that a randomly selected adult has an IQ greater than 134.1.

This is 1 subtracted by the pvalue of Z when X = 134.1. So

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

`Z = (134.1 - 102.6)/(23.6)`

`Z = 1.33`

`Z = 1.33` has a pvalue of 0.9082.

1-0.9082 = 0.0918

So there is a 9.18% probability that a randomly selected adult has an IQ greater than 134.1