Respuesta :
Using the z-distribution, it is found that the 90% confidence interval for the population mean is: 132.10 to 143.90.
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 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.
The other parameters are given as follows:
[tex]\mu = 138, \sigma = 34, n = 90[/tex]
Hence, the bounds of the interval are given by:
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 138 - 1.645\frac{34}{\sqrt{90}} = 132.1[/tex]
[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 138 + 1.645\frac{34}{\sqrt{90}} = 143.9[/tex]
The 90% confidence interval for the population mean is: 132.10 to 143.90.
More can be learned about the z-distribution at https://brainly.com/question/25890103
#SPJ5
