There is no single magic sample size that guarantees that sampling distributions will be approximately normal. If a population is normal, a sample size of one suffices. If a population is fairly symmetrical, small sample sizes usually suffice. For strongly skewed populations, n may need to be quite large for the sampling distribution to be approximately normal. For many distributions that arise in practice, relativley small sample sizes such as 20 or 30 are sufficiently large for the normal approximation to hold. Do know, however, that there are exceptions.
Problem: Birth weights have a mean of 120 ounces and a standard deviation of 21 ounces. What is the probability that the average birth weight of twenty new babies is smaller than 105 ounces?
Solution: By the CLT, the sampling distribution is approximately normal. (We are assuming that birthweights are not highly skewed.) The z-score is
z = (105 - 120) / (21 / sqrt(20)) = -3.19.Notice that we use the standard error and not the population standard deviation in the denominator because the problem asks about a sample mean, not a single individual. The area to the left of -3.19 under a standard normal curve is 0.0007. It is rather unlikely for a sample mean of twenty birth weights from this population to be this small.
Problem: Find the middle 90% of sample mean birth weights from the previous problem for a sample size of 20.
Solution: The lower and upper limits of the middle 90 percent are the 5th and 95th percentiles. The z-score that corresponds to the 95th percentile is 1.645. Thus, the middle 90 percent of mean birth weights from samples of size 20 fall between
120 - 1.645(21)/sqrt(20) and 120 + 1.645(21)/sqrt(20) or 112.3 and 127.7
Bret Larget, email@example.com