Question

Please help! I don't understand how to solve this problem

Answer 1

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

What is a z-distribution confidence interval?

The confidence interval is:

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

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

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.

Assuming an uniform distribution, the standard deviation is given by:

$S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547$

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.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

$M = z\frac{\sigma}{\sqrt{n}}$

$0.006 = 1.96\frac{1.1547}{\sqrt{n}}$

$0.006\sqrt{n} = 1.96 \times 1.1547$

$\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}$

$(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2$

n =  142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

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

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