Question

A population has a mean, μ = 89 and a standard deviation,σ = 24.

Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 64.

Answer 1

`\bar X \sim N(\mu=89, (24)/(√(64))=3)`

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".

Solution to the problem

Let X the random variable of interest for a population. We know from the problem that the distribution for the random variable X is given by:

`X\sim N(\mu =89,\sigma =24)`

We take a sample of n=64 . That represent the sample size.

The sample mean is defined as:

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

And if we find the expected value and variance for the sample mean we got:

`E(\bar X) = \mu`

Var(\bar X) = \frac{\sigma^2}{n}[/tex]

The distribution for the sample mean is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

`\bar X \sim N(\mu=89, (24)/(√(64))=3)`