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

#SPJ1

Otras preguntas