Question

A random variable x 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 x ?

Answer 1

Answer:

95% Confidence Interval: (94.08,101.92)    

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = 98

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = 12

95% Confidence Interval:

Formula:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

$98 \pm 1.96(\dfrac{12}{\sqrt{36}} ) = 98 \pm 3.92 = (94.08,101.92)$