Question

A random sample of n D 225 and xN D 21 was drawn from a normal population with a known

Answer 1

Using the z-distribution, the 95% confidence interval of the population mean is: (17.5, 24.5).

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so the critical value is z = 1.96.

For this problem, the other parameters are:

$\overline{x} = 21, \sigma = 26.8, n = 225$

Hence the bounds of the interval are:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 21 - 1.96\frac{26.8}{\sqrt{225}} = 17.5$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 21 + 1.96\frac{26.8}{\sqrt{225}} = 24.5$

The interval is: (17.5, 24.5).

More can be learned about the z-distribution at brainly.com/question/25890103

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