Answer: The 95% confidence interval for the mean of x is (94.08, 101.92) .
Step-by-step explanation:
We are given that ,
A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.
i.e. [tex]\sigma= 12[/tex]
Also, it is given that , Sample mean [tex]\overline{x}=98[/tex] having sample size : n= 36
For 95% confidence ,
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
By using the z-value table , the two-tailed critical value for 95% Confidence interval :
[tex]z_{\alpha/2}=z_{0.025}=1.96[/tex]
We know that the confidence interval for unknown population mean[tex](\mu)[/tex] is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\sigma[/tex] = Population standard deviation
[tex]z_{\alpha/2}[/tex] = Critical z-value.
Substitute all the given values, then the required confidence interval will be :
[tex]98\pm (1.96)\dfrac{12}{\sqrt{36}}[/tex]
[tex]=98\pm (1.96)\dfrac{12}{6}[/tex]
[tex]=98\pm (1.96)(2)[/tex]
[tex]=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)[/tex]
Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .
