Question

IQ scores are normally normally distributed with a mean of 100 and center deviation of 15 if one person is randomly selected what is the probability that the persons IQ far between 110 and 130

Answer 1

22.86% probability that the persons IQ is between 110 and 130

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 = 100, \sigma = 15`

If one person is randomly selected what is the probability that the persons IQ is between 110 and 130

This is the pvalue of Z when X = 130 subtracted by the pvalue of Z when X = 110.

X = 130

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

`Z = (130 - 100)/(15)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

X = 110

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

`Z = (110 - 100)/(15)`

`Z = 0.67`

`Z = 0.67` has a pvalue of 0.7486

0.9772 - 0.7486 = 0.2286

22.86% probability that the persons IQ is between 110 and 130