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

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(\sigma)/(√(n))\\\\=98\pm 1.96(12)/(√(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).

Answer 2

Answer: (94.08, 101.92)

Explanation:

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

`\overline{x}\pm z^*(\sigma)/(√(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)(12)/(√(36))`

`=98\pm (1.96)(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) .