Sampling Distributions Scores on the Gilliam Autism Rating Scale (GARS) are normally distributed with a mean of 100 and a standard deviation of 15. A sample of 64 GARS scores is randomly selected and the sample mean is computed.

a. Describe the distribution of such sample means.

b. What is the mean of all such sample means?

c. What is the standard deviation of all such sample means?

It is given that the scores on the Gilliam Autism Rating Scale (GARS) are normally distributed with a mean of 100 and a standard deviation of 15.

A random sample of size 64 is selected from this population.

a.

According to the central limit theorem, when a population follows the normal distribution and samples are selected from this population, then the sampling distribution of the sample means also follows the normal distribution.

Here, the population of scores is normally distributed, and a sample is selected whose mean value is computed.

Thus, the sampling distribution of all such sample mean scores is normally distributed.

b.

The mean value of all sample means is equal to the population mean.

Mathematically,

\({\mu _{\bar x}} = \mu \)where

\({\mu _{\bar x}}\)is the mean of all sample means

\(\mu \)is the population mean.

Here, the population mean score is given to be equal to\(\mu = 100\).

Therefore, the mean of all sample mean scores is equal to 100.

c.

The standard deviation of all sample means has the following formula:

\({\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }}\)where

\(\sigma \)is the population standard deviation

n is the sample size

Here, the population standard deviation is equal to 15, and the sample size is equal to 64.

Thus, the value of the standard deviation of sample values is computed as follows:

\(\begin{aligned}{c}{\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }}\\ = \frac{{15}}{{\sqrt {64} }}\\ = \frac{{15}}{8}\\ = 1.875\end{aligned}\)

Therefore, the standard deviation of all sample means is equal to 1.875.