Question

IQ test scores are normally distributed with a mean of 100 and a standard deviation of 15.

(a) Find the IQ scores that represent the bottom 35%.
(b) Find the IQ score that represents the 3rd Quartile.
(c) Find the IQ score for the top 5%.

Answer 1

a) IQ scores of 94.2 and below represent the bottom 35%.

b) An IQ score of 110.1 represents the 3rd quartile.

c) IQ scores of 124.7 and higher are in the top 5%.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 100 and a standard deviation of 15.

This means that
`\mu = 100, \sigma = 15`

(a) Find the IQ scores that represent the bottom 35%.

The 35th percentile and below, in which the 35th percentile is X when Z has a p-value of 0.35, so X when Z = -0.385.

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

`-0.385 = (X - 100)/(15)`

`X - 100 = -0.385*15`

`X = 94.2`

IQ scores of 94.2 and below represent the bottom 35%.

(b) Find the IQ score that represents the 3rd Quartile.

This is the 100*3/4 = 75th percentile, which is X when Z has a p-value of 0.75, so X when Z = 0.675.

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

`0.675 = (X - 100)/(15)`

`X - 100 = 0.675*15`

`X = 110.1`

An IQ score of 110.1 represents the 3rd quartile.

(c) Find the IQ score for the top 5%.

IQ scores of at least the 100 - 5 = 95th percentile, which is X when Z has a p-value of 0.95, so X when Z = 1.645.

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

`1.645 = (X - 100)/(15)`

`X - 100 = 1.645*15`

`X = 124.7`

IQ scores of 124.7 and higher are in the top 5%.