In: Statistics and Probability
. Assume X ~ N (10, 4).
Given
i. The approximate distribution of sample mean \bar{X} if the
sample size n = 100,
is Normally distributed
According the Central Limit Theorem, consider an iid
sample 

Then as the sample size increases, the sample mean converges in
distribution to a standard normal

So the approximate distribution of \bar{X} for a sample size n is


In our case with

ii. As sample size increases, the variance of \bar{X} decreases


So as sample size n increases, variance decreases
Here is the sampling distribution of \bar{X} as a function of
n,
You can see in the plot below that the curve becomes less wider as
n increases,
hence the variance decreases

iii) The two values (symmetric around the population mean) that
contain
of the population can be derived as follows:
Let L and U be the standardized lower and upper bounds of the interval, then as it is a symmetric interval,


\implies


(This is the symmetric interval)

The plot of the symmetric interval is given below

Hence for
a) 75% interval,


Now the interval is given by,

In our case with

a) 95% interval,


Now the interval is given by,

In our case with
