Question

Data from a sample of 576 people in a city was used to estimate the mean monthly bill

Answer 1

Using the z-distribution, it is found that the margin of error for a 95% confidence interval is of $0.04.

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 sample.

The margin of error is given by:

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

In this problem, the values of the parameters are:

$z = 1.96, \sigma = 0.48, n = 576$.

Hence, the margin of error, in dollars, is given by:

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

$M = 1.96\frac{0.48}{\sqrt{576}}$

M = 0.04.

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