Respuesta :
Using the z-distribution, the 99% confidence interval of the population mean number of days it takes to find a job is (35.38, 44.62).
What is a z-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm 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.
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.
The other parameters are:
[tex]\overline{x} = 40, \sigma = 10, n = 31[/tex]
Hence the bounds of the interval are:
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 40 - 2.575\frac{10}{\sqrt{31}} = 35.38[/tex]
[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 40 + 2.575\frac{10}{\sqrt{31}} = 44.62[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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