Question

In a large population of adults, the mean IQ is 112 with a standard deviation of 20. Suppose 200 adults are randomly selected for a market research campaign. The distribution of the sample mean IQ is

Answer 1

`X \sim N(112,20)`

Where
`\mu=112` and
`\sigma=20`

Since th distribution for X is normal then the distribution for the sample mean
`\bar X` is also normal and given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

`\mu_(\bar x) = 112`

`\sigma_(\bar x)= (20)/(√(200))=1.414`

`\bar X \sim N( 112,1.414)`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(112,20)`

Where
`\mu=112` and
`\sigma=20`

Since th distribution for X is normal then the distribution for the sample mean
`\bar X` is also normal and given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

`\mu_(\bar x) = 112`

`\sigma_(\bar x)= (20)/(√(200))=1.414`

`\bar X \sim N( 112,1.414)`