Question

Assume adults have IQ scores that are normally distributed with a mean of 102 and standard deviation of 16. Find the probability that a randomly selected individual has an IQ between 81 and 109

Answer 1

57.49% probability that a randomly selected individual has an IQ between 81 and 109

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 = 102, \sigma = 16`

Find the probability that a randomly selected individual has an IQ between 81 and 109

This is the pvalue of Z when X = 109 subtracted by the pvalue of Z when X = 81. So

X = 109

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

`Z = (109 - 102)/(16)`

`Z = 0.44`

`Z = 0.44` has a pvalue of 0.67

X = 81

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

`Z = (81 - 102)/(16)`

`Z = -1.31`

`Z = -1.31` has a pvalue of 0.0951

0.67 - 0.0951 = 0.5749

57.49% probability that a randomly selected individual has an IQ between 81 and 109