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

Answer:

$X \sim N(100,15)$  

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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Because by definition:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$E(\bar X) = \mu$

$Var(\bar X) = \frac{\sigma^2}{n}$

And for this case we have this:

$\mu_{\bar X}= \mu = 100$

$\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75$

Step-by-step 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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Because by definition:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$E(\bar X) = \mu$

$Var(\bar X) = \frac{\sigma^2}{n}$

And for this case we have this:

$\mu_{\bar X}= \mu = 100$

$\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75$