Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What percentage of people have an IQ score between 69 and 132, to the nearest tenth?

Answer 1

96.4% of people have an IQ score between 69 and 132

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:

`\mu = 100, \sigma = 15`

What percentage of people have an IQ score between 69 and 132, to the nearest tenth?

The proportion is the pvalue of Z when X = 142 subtracted by the pvalue of Z when X = 69. The percentage is the proportion multiplied by 100. So

X = 132

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

`Z = (132 - 100)/(15)`

`Z = 2.13`

`Z = 2.13` has a pvalue of 0.983

X = 69

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

`Z = (69 - 100)/(15)`

`Z = -2.07`

`Z = -2.07` has a pvalue of 0.019

0.983 - 0.019 = 0.964

0.964*100 = 96.4%

96.4% of people have an IQ score between 69 and 132