Question

you have information to suggest that a certain continuous variable of a population has a mean of \\( \mu = 13.24 \\) and a standard deviation of \\( \sigma = 5.12 \\). you are to randomly pick \\( n = 54 \\) individuals from this population and observe the value of the population variable on each. the value is to measured as \\( x \\).
after the random sample of \\( n = 54 \\) has been taken, you are asked to consider the behavior of the statistic \\( \overline{x} \\).
(a) complete the statement below. use as many decimals as you can.
the distribution of \\( \overline{x} \\) with a mean \\( \mu_{\overline{x}} = \\) and a standard deviation \\( \sigma_{\overline{x}} = \\)
(b) find the probabilit is approximately normal \\( n = 54 \\) is between 12.64 and 13.74. use at least four decimals in your z - values and enter your answer to at least four decimals. is exactly normal is skewed to the right has an unknown shape

Explanation:

Step1: Mean of the sample mean

By the property of the sample mean, the mean of the sampling distribution of \(\bar{X}\) is equal to the population mean. So, \(\mu_{\bar{X}}=\mu = 13.24\).

Step2: Standard deviation of the sample mean

The formula for the standard deviation of the sampling distribution of \(\bar{X}\) (also known as the standard error) is \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 3.12\) and \(n = 54\), we have \(\sigma_{\bar{X}}=\frac{3.12}{\sqrt{54}}\approx\frac{3.12}{7.3485}\approx0.4246\).

Answer:

The distribution of \(\bar{X}\) is approximately Normal (by the Central Limit Theorem since \(n = 54\geq30\)) with a mean \(\mu_{\bar{X}} = 13.24\) and a standard deviation \(\sigma_{\bar{X}}\approx0.4246\).