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 ?

Respuesta :

Answer:

95% Confidence Interval: (94.08,101.92)    

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 98

Sample size, n = 36

Alpha, α = 0.05

Population standard deviation, σ = 12

95% Confidence Interval:

Formula:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.05} = 1.96[/tex]

[tex]98 \pm 1.96(\dfrac{12}{\sqrt{36}} ) = 98 \pm 3.92 = (94.08,101.92)[/tex]