# Central Limit Theorem and Skewed Distribution

I'm looking for a simple answer to this question relating the central limit theorem and Gaussian and skewed distributions, if one exists. I used the binomial function to generate calculations of the probabilities of possible outcomes for 10 flips of an unfair coin (p=0.3, q=0.7) and obtained a skewed distribution. I have been thinking of this as a kind of sampling distribution of proportions. If the coin was flipped 10 trillion times, with 3 trillion heads and 7 trillion tails, and 10 flip samples are plotted into the sampling distribution, yielding my skewed curve.

Now I also "learned" that the central limit theorem says that the sampling distribution of any distribution is a Gaussian curve, but I acknowledge that that my studies of this are relatively superficial. Is a skewed curve still considered a Gaussian curve? Are there other important aspects about the central limit theorem that I am clearly unaware of? I'm not looking for a comprehensive explanation necessarily, but just some guidance about misconceptions that I may have.

Thanks.

• (1) The CLT does not apply to all distributions; it is required that the variance exist. (2) Even if the IID random variables being averaged are skewed, the CLT still applies and says that the average tends to (symmetrical) normal. – BruceET Dec 4 '20 at 21:26
• The central limit theorem relates to a limit distribution and not to the sample distribution of the sum of 10 coin flips. This question has been asked before here. – Sextus Empiricus Dec 4 '20 at 21:49
• Of possible interest: stats.stackexchange.com/questions/473455/… – Dave Dec 4 '20 at 22:03
• stats.stackexchange.com/questions/389590/… – Sextus Empiricus Dec 4 '20 at 22:03
• The first portion of my post on the CLT aims at heading off many common misconceptions. – whuber Dec 4 '20 at 22:08

Suppose $$X_i \stackrel{iid}{\sim} \mathsf{Binom}(n = 10, p=0.3),$$ a skewed distribution.

plot(x, PDF, type="h", lwd=3, col="blue",
main="PDF of BINOM(10, .3)")
abline(h=0, col="green2")
abline(v=0, col="green2")


Then the average $$\hat p = \bar X_{1000}$$ of $$m = 1000$$ of these $$X_i$$s is very nearly normal, as illustrated in the following simulation in R, based on 100,000 replications of this estimate $$\hat p$$ for $$p.$$

set.seed(2020)
p.est = replicate(10^5, mean(rbinom(1000, 10, .3)))
summary(p.est)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
2.814   2.969   3.000   3.000   3.031   3.195
sd(p.est)
[1] 0.04594069

hist(p.est, prob=T, col="skyblue2",
main="Simulated Sampling Dist'n")
col="orange", lwd=2)


According to a Shapiro-Wilk test on the first 5000 simulated values of $$\hat p,$$ they are consistent with a random sample from a normal distribution. [The S-W test in R is restricted to a maximum of 5000 observations.]

shapiro.test(p.est[0:5000])

Shapiro-Wilk normality test

data:  p.est[0:5000]
W = 0.99971, p-value = 0.727


Nevertheless, The distribution of $$\hat p$$s based on a thousand observations is discrete (even though the histogram doesn't reveal that). Among the 100,000 realizations of $$\hat p$$ from the simulation above, there are only 355 unique values.

length(unique(p.est))
[1] 355


This is to follow up from my initial question. I calculated frequency distributions for n = 10, 25, and 100 for an even more unfair coin (p=0.1, q=0.9) and found, by eye at least, that the data became much more symmetrical at higher numbers of coin flips. My takeaway is that the binomial function calculated frequency distribution is a sampling distribution of a Bernoulli type experiment with a global population of trillions of 0's and 1's in frequency ratio p and q, selected with individual sample size n. As n increases, the distribution appears to become increasingly symmetrical bell shaped.

I don't know however if the data becomes more normal per the Shapiro-Wilk normality test mentioned in BruceET's answer. I calculated the expected frequencies on a spreadsheet, and suspect that the data points will not meet SW criteria even though the curves look bell shaped. I think this because as a check of how the SW normality test works on statskingdom, I manually entered 1024 data points for the expected sampling distribution frequencies of flipping a fair coin 10 times, and webpage result was that the binomial distribution was not Gaussian. Perhaps it would have been different with a larger set of points from a binomial distribution.