Arandom 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 ?

(94.08, 101.92)

Step-by-step explanation:

The confidence interval for unknown population mean is given by :-

, where = Sample mean

= 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.

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 :-

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

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like: