Respuesta :
Using the z-distribution, considering [tex]\overline{x} = 36, \sigma = 7[/tex], the 95% confidence interval for the mean is (32.67, 39.33).
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 sample.
We have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
The other parameters, researching the problem on the internet, are given as follows:
[tex]\mu = 36, \sigma = 7, n = 17[/tex].
Hence the bounds of the interval are:
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 36 - 1.96\frac{7}{\sqrt{17}} = 32.67[/tex]
[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 36 + 1.96\frac{7}{\sqrt{17}} = 39.33[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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