Using the z-distribution, it is found that the margin of error of the confidence interval is:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, it is stated that the population standard deviation is known, thus, the z-distribution is used.
The margin of error of a z-confidence interval is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
A similar problem is given at https://brainly.com/question/25214883