Respuesta :
Using the z-distribution, as we have the population standard deviation, it is found that a sample of 96 people must be taken.
What is the margin of error for the z-distribution?
It is given by:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- z is the critical value.
- [tex]\sigma[/tex] is te population standard deviation.
- n is the sample size.
In this problem, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
Additionally, we have that [tex]\sigma = 38, M = 10[/tex], hence we solve for n to find the minimum sample size needed.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]10 = 2.575\frac{38}{\sqrt{n}}[/tex]
[tex]10\sqrt{n} = 38 \times 2.575[/tex]
[tex]\sqrt{n} = 3.8 \times 2.575[/tex]
[tex](\sqrt{n})^2 = (3.8 \times 2.575)^2[/tex]
n = 95.7
Rounding up, a sample of 96 people must be taken.
More can be learned about the z-distribution at https://brainly.com/question/25890103
