Question

A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a random sample of size n = 36 and find a sample mean of ¯ x = 98 . What is a 95% confidence interval for the mean of x ?

Answer 1

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. $\sigma= 12$

Also, it is given that , Sample mean $\overline{x}=98$ having sample size : n= 36

For 95% confidence ,

Significance level : $\alpha=1-0.95=0.05$

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

$z_{\alpha/2}=z_{0.025}=1.96$

We know that the confidence interval for unknown population mean$(\mu)$ is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

$\sigma$ = Population standard deviation

$z_{\alpha/2}$ = Critical z-value.

Substitute all the given values, then the required confidence interval will be :

$98\pm (1.96)\dfrac{12}{\sqrt{36}}$

$=98\pm (1.96)\dfrac{12}{6}$

$=98\pm (1.96)(2)$

$=98\pm 3.92=(98-3.92,\ 98+3.92)\\\\=( 94.08,\ 101.92)$

Therefore, the 95% confidence interval for the mean of x is (94.08, 101.92) .