# If the parent distribution is skewed, but the sample size is greater than 30, is the distribution of the sample mean nearly Normal?

I have my AP Stats exam tomorrow and had this question.

I apologize if it is too simplistic for this forum.

I'm pretty sure the sample's distribution is nearly normal, but not 100%, can someone enlighten me?

Thank you!

• You need to know that the requirements for the central limit theorem hold. That includes the existence of a variance. Let's assume that the central limit theorem hold you still need to know something about the rate of convergence to the normal. The number 30 is used because for a wide range of values for the success parameter p of a Binomial distribution 30 is an adequate sample size for the sample estimate of p to have an approximate Normal distribution. May 16 '18 at 18:11
• See stats.stackexchange.com/questions/69898/… for examples where the sampling distribution is decidedly non-normal.
– whuber
May 16 '18 at 18:17
• Strong overlap with this question: stats.stackexchange.com/questions/61798/… ... not quite sure that it counts as a duplicate May 17 '18 at 0:12

The answer to the question in your title is "no" (sometimes it would be the case - if you pin down what you mean by nearly and define the circumstances, it may happen - but the number 30 is essentially irrelevant in general; it's no more special than 10 or 100 or 100000).

For an example, consider a skewed parent distribution - say the gamma distribution with shape parameter 0.01 and whatever (common) scale you like. The mean of 30 of them is also-gamma now with shape 0.3 (and a different scale). It's still quite skew and clearly non-normal. At n=100, you get an exponential distribution for the mean -- still pretty skew, still non-normal. At n=3000 ... it's starting to look pretty reasonable and for some definitions of "nearly" would be fine (but if you're trying to work out extreme tail probabilities, it may still not be enough).

[The mere fact that they mention n=30 has me concerned that they expect an answer which will be wrong. Check your materials/notes/text to be sure that you know what answer they expect to get, irrespective of the actual situation (do I bemoan the need to give this advice? Why, yes, indeed I do).]

As other answers have pointed out, convergence to the normal distribution depends on whether the conditions for the CLT apply (i.e., whether the parent distribution falls within the conditions for the theorem$^\dagger$) and the "nearness" of this convergence for a particular value of $n$ is something that can be measured in various ways. Hence, the answer to your question really depends on what your examiner means by the "nearly" normal (i.e., how near is "nearly").

Questions like this are inherently vague, since they depend on how near two distributions have to be to say that one is "nearly" the other. This means that you have a bit of latitude here in describing the "nearness" of the distribution to normality. Probably the best way to understand and describe the rate of convergence to normality under the CLT is just to go through some examples, using a non-normal parent distribution, and see what the distribution of the sample mean looks like for different values of $n$. Do this by starting with a tractable parent distribution (e.g., gamma mixtures) and deriving the distribution of the sample mean for a value of $n$, and then have a look at it.

Skewness of the sample mean: Since your question is asking about skewness in the parent distribution, it is useful to concentrate on this aspect of the distribution. Skewness in the parent distribution flows through to skewness in the distribution of the sample mean, but the averaging process in the sample mean leads to a reduction in the skewness as you get more and more data.

Fortunately, the skewness of the distribution of the sample mean for IID data can be obtained without use of the CLT, so it can be obtained for any parent distribution. If the parent distribution has skewness $\mathbb{Skew}(X) = \gamma$ then you have $\mathbb{Skew}(\bar{X}_n) = \gamma / \sqrt{n}$. With $n=30$ you get:

$$\mathbb{Skew}(\bar{X}_{30}) = \frac{1}{\sqrt{30}} \cdot \mathbb{Skew}(X) = 0.1825742 \cdot \mathbb{Skew}(X).$$

So you can see that $n = 30$ data points gives you a skewness that is about 18% of the skewness of the parent distribution, which is a pretty good reduction. For a very strongly skewed parent distribution this might still be a high skewness, but for weakly skewed parent distributions, this is probably not much skew. This result does not depend on the CLT, so you needn't ask for any more conditions. However, broader convergence to normality does require the conditions of the CLT to be fulfilled, so that is worth mentioning also.

$^\dagger$ The classical CLT requires IID data from a parent distribution with finite variance. This rules out parent distributions whose tails decrease too slowly to yield a finite variance. These "heavy-tailed" distributions have different convergence results, and the distribution of the sample mean in these cases is not necessarily asymptotically normal.

By the central limit theorem, the sampling distribution of the mean tends towards Gaussian regardless of the form of the parent (though there are some exceptions such as Cauchy). More so as the sample size or number of samples increases. There are some nice simulators online that demonstrate this visually.

• There are lots of exceptions. Any distribution without a variance and all the distributions that have the mean converge to a stable law. May 17 '18 at 0:26