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 n=36 and find a sample mean of ¯ x = 98 x¯=98 . What is a 95% confidence interval for the mean of x ?

Answer 1

Answer: (94.08, 101.92)

Step-by-step explanation:

The confidence interval for unknown population mean$(\mu)$ is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}$

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

$\sigma$ = Population standard deviation

z* = Critical z-value.

Given : A random variable x has a Normal distribution with an unknown mean and a standard deviation of 12.

$\sigma= 12$

$\overline{x}=98$

n= 36

Confidence interval = 95%

We know that the critical value for 95% Confidence interval : z*=1.96

Then, the 95% confidence interval for the mean of x  will be :-

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

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

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

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

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

Answer 2

Answer:  95% confidence interval would be (94.08,101.92).

Step-by-step explanation:

Since we have given that

n = 36

standard deviation = 12

sample mean = 98

At 95% confidence, z = 1.96

So, interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=98\pm 1.96\dfrac{12}{\sqrt{36}}\\\\=98\pm 3.92\\\\=(98-3.92,98+3.92)\\\\=(94.08,101.92)$

Hence, 95% confidence interval would be (94.08,101.92).