# Distribution of the average

If the true population average is 15 and we are going to take a random sample and calculate the average 1,000,000 times, what is the distribution of the estimated average?

My thoughts:

By the CLT, $$\frac{\bar{x} - E\bar{x}}{\sqrt{Var {\bar{x}}}} \sim Normal(0,1)$$ as the number of trials to calculate the mean approaches infinity. So, the distribution of the estimated average should be $$Normal(15, Var(\bar{x}))$$, but $$Var(\bar{x})= Var(X)/n$$, where $$X$$ is a random number from the population and $$n$$ is the size of the random sample.

Is this right? So, are the random samples should be of the same size n?

The Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution as the sample size approaches infinity.

If you have a fixed sample size, then the sampling distribution will not actually be normal in most cases - for a very simple example, if the sample size is 1 then the distribution of the sample mean is the distribution of the population (because all you're doing is measuring one value from the population at a time).

However, if the sample mean is "big enough", then you can indeed say that the distribution of the sample mean is approximately normal, with the distribution you say, assuming that the sample is of fixed size. As it stands, the question seems to be missing some key information to comment more accurately.

• I am not sure. They are saying that you get a sample a million times. So, I think they take a sample, get the mean, then repeat the process. This is how they get a sample of means. Am I right regarding the understanding of the question? The question is "what probability distribution would the estimated averages approximate?"
– Vika
Commented May 6, 2020 at 23:25
• I had the same question about the exact process described in the question, but I see nothing wrong with what is said in this answer. (+1) Commented May 7, 2020 at 5:34
• &BruceET So, as a previous commenter said, if the size of each sample is 1, but we do sampling 1 million times and thus get 1 million averages, they will still follow normal distribution, right?
– Vika
Commented May 7, 2020 at 17:24
• &ConMan When you said "the sample mean is "big enough", did you mean the sample size "big enough"?
– Vika
Commented May 7, 2020 at 17:25
• I did mean sample size, my mistake. And if you draw a million samples, each of size 1, then you're just drawing a million values from the population - which follows the distribution of the population itself. So if the population has, say, a uniform distribution, then you'd expect the million samples to be uniform too. Commented May 7, 2020 at 23:02

Comment: I'm wondering (a) if the following shows what you're doing, and (b) What is the purpose of this? A classroom demonstration on the CLT? Or something else?

In R, we simulate a million realizations of $$A = \bar X,$$ the sample average of $$n = 900$$ values from $$\mathsf{Norm}(\mu = 15, \sigma = 3).$$

set.seed(506)
a = replicate( 10^6, mean(rnorm(900, 15, 3)) )
mean(a)
[1] 14.99992     # aprx E(samp avg) = 15
var(a)
[1] 0.009992273  # aprx V(samp avg) = 9/900 = 0.01


Histogram of the one million sample averages $$A = \bar X.$$ The red curve is the density of $$\mathsf{Norm}(\mu = 15, \sigma = 1/10).$$

hist(a, prob=T, br=40, col="skyblue2")


Again, but with data sampled from a (right-skewed) exponential distribution. [R uses rate parameter $$\lambda = 1/\mu]:$$

set.seed(2010)
a = replicate( 10^6, mean(rexp(900, .1)) )
mean(a)
[1] 10.00039     # aprx E(samp mean) = 10
var(a)
[1] 0.1112148    # aprx Var(samp mean) = 100/900 = 0.1111


This time the distribution of $$A = \bar X$$ is very nearly normal, but not exactly (still a tiny bit of skewness, hardly visible in plot). The red curve is the density of $$\mathsf{Norm}(\mu = 10, \sigma = 1/3);$$ the (almost coincident) black dotted curve is the density curve of the exact distribution $$\mathsf{Gamma}(\mathrm{shape}=900, \mathrm{rate} = 90)$$ of $$A = \bar X.$$

hist(a, prob=T, br=40, col="skyblue2")

• Good: If data are normal, then $n$ doesn't matter because the sample mean will be normal also, regardless of $n.$ Then the issue is to show that $Var(\bar X) = \sigma^2/n.$ If data are not normal, then $n$ should be large enough that the dist'n of $\bar X$ is nearly normal. (For exponential data, $n=900$ seems large enough. Commented May 7, 2020 at 19:19