# Central Limit Theorem - Significance of Sample Count

According to a Khan Academy's lecture, the Central Limit Theorem is defined as follows:

Central limit theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution.

To test this definition I considered a population of 100,000 random numbers with the following parameters (see the image below)

Population Parameters:

Mean: 503.76, Median: 503.0, Mode: 338, and Standard Deviation: 285.72

Then, plotting the sampling distribution of the sample mean with varying sample sizes and sample counts resulted in the following observations (each graph is accordingly labeled).

Question: It seems to me that simply increasing the sample size is not sufficient for the distribution to become normal (based on visual observation). The number of samples also should be more.

1. Then how do I reconcile these observations with the formal definition of the theorem?
2. And are these conclusions correct (given the plots)?
• Increasing the sample size simply reduces the standard error
• As long as the sample size is above a minimum value, increasing the sample count seems sufficient for normality
• I'd adapt the x-axis range to the standard error for making statements about how normal it looks. Currently you can't really see much of the distributional shape in the n=1000 plot. It doesn't look all too non-normal to me, but I can't tell either if everything is concentrated in the middle. Jun 30 '19 at 21:17
• The statement made in the Khan quote is inaccurate; if the variance is finite (and some other conditions hold), asymptotically the distribution of the sample mean goes to a constant, the population mean (see the law of large numbers). [However, the standardized sample mean $\frac{\bar{X}_n-\mu}{\sigma/\sqrt{n}}$ will go to a standard normal, or equivalently, $\sqrt{n}(\bar{X}_n-\mu)$ will go to a normal with mean $0$ and variance $\sigma^2$. This may impact how you investigate progress toward it.] Jun 30 '19 at 23:09
• @Lewian I plotted the graphs by adapting the x-axis range to the range of distribution and the plots still look normal. I think a better estimate of normality would be obtained through a Q-Q plot and/or by calculating Kurtosis and Skew. Jul 2 '19 at 5:09

Just for remaining, this is the central limit theorem :

Suppose $$\{X_1, X_2, …,X_n\}$$ is a sequence of i.i.d. random variables with $$\mathbb{E}[X_i] = \mu$$ and $$\mathbb{V}ar[X_i] = \sigma^2 < \infty$$. Then as $$n$$ approaches infinity, the random variables $$\sqrt{n}(\overline{X}_n − \mu)$$ where $$\overline{X}_n = \frac{\sum_i X_i}{n}$$ converge in distribution to a normal $$\mathcal{N}(0,\sigma^2)$$ :

$$\displaystyle {\sqrt {n}}\left(\overline{X}_n-\mu \right)\ {\xrightarrow {d}}\ N\left(0,\sigma ^{2}\right).$$

We can also say $$\displaystyle \frac{\left(\overline{X}_n-\mu \right)}{\sigma/\sqrt {n}}\ {\xrightarrow {d}}\ N\left(0,1\right).$$

We can see that $$\mathbb{E}[\overline{X}_n] = \mu$$ and $$\mathbb{V}ar[\overline{X}_n] = \frac{\sigma^2}{\sqrt{n}}$$.

First, it is $$\overline{X}_n$$ which is a random variable and can be normal distributed. To plot a distibution of $$\overline{X}_n$$ you need many realizations or observations of this variable. Then you need many "sample count". By increasing $$n$$, you still have only one value (realization of the variable). It just help you to be closer to the real value of $$\mu$$ due to law of large number but not to plot the law of this variable.
Second, the variance of $$\overline{X}_n$$ is $$\frac{\sigma^2}{\sqrt{n}}$$ with help you to see that as $$n$$ increase, the variance converge to $$0$$.