Respuesta :
Using the z-distribution, it is found that a sample of 171 should be selected.
What is a z-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
The margin of error is:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- z is the critical value.
- n is the sample size.
- [tex]\sigma[/tex] is the standard deviation for the population.
For this problem, the parameters are:
[tex]z = 1.96, \sigma = 40, M = 6[/tex]
Hence we solve for n to find the needed sample size.
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]6 = 1.96\frac{40}{\sqrt{n}}[/tex]
[tex]6\sqrt{n} = 40 \times 1.96[/tex]
[tex]\sqrt{n} = \frac{40 \times 1.96}{6}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{40 \times 1.96}{6}\right)^2[/tex]
n = 170.7.
Rounding up, a sample of 171 should be selected.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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