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

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

Step-by-step explanation:

We are given that ,

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

i.e.

Also, it is given that , Sample mean having sample size : n= 36

For 95% confidence ,

Significance level :

By using the z-value table , the two-tailed critical value for 95% Confidence interval :

We know that the confidence interval for unknown population mean is given by :-

, where = Sample mean

= Population standard deviation

= Critical z-value.

Substitute all the given values, then the required confidence interval will be :

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

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

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step-by-step explanation: