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

Answer:

$\bar X \sim N(\mu=89, \frac{24}{\sqrt{64}}=3)$

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".  

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:

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

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:

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

$\bar X \sim N(\mu=89, \frac{24}{\sqrt{64}}=3)$