Question

It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains a sample of 16 students’ IQ scores from a statistics class at UT. What is the shape of this sampling distribution?

Answer 1

`X \sim N(100,15)`

Where
`\mu=100` and
`\sigma=15`

We select a sample of n=16 and we are interested on the distribution of
`\bar X`, since the distribution for X is normal then we can conclude that the distribution for
`\bar X` is also normal and given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

Because by definition:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

`E(\bar X) = \mu`

`Var(\bar X) = (\sigma^2)/(n)`

And for this case we have this:

`\mu_(\bar X)= \mu = 100`

`\sigma_(\bar X) = (15)/(√(16))= 3.75`

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(100,15)`

Where
`\mu=100` and
`\sigma=15`

We select a sample of n=16 and we are interested on the distribution of
`\bar X`, since the distribution for X is normal then we can conclude that the distribution for
`\bar X` is also normal and given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

Because by definition:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

`E(\bar X) = \mu`

`Var(\bar X) = (\sigma^2)/(n)`

And for this case we have this:

`\mu_(\bar X)= \mu = 100`

`\sigma_(\bar X) = (15)/(√(16))= 3.75`