Question

Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105μ=105 and a standard deviation sigma equals 20σ=20. Find the probability that a randomly selected adult has an IQ between 8989 and 121121.

Answer 1

57.62% probability that a randomly selected adult has an IQ between 89 and 121.

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 = 105, \sigma = 20`

Find the probability that a randomly selected adult has an IQ between 89 and 121.

This is the pvalue of Z when X = 121 subtracted by the pvalue of Z when X = 89. So

X = 121

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

`Z = (121 - 105)/(20)`

`Z = 0.8`

`Z = 0.8` has a pvalue of 0.7881

X = 89

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

`Z = (89 - 105)/(20)`

`Z = -0.8`

`Z = -0.8` has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

57.62% probability that a randomly selected adult has an IQ between 89 and 121.