### Section 5.3: Distribution of the Sample Mean

#### Key Concepts

The sampling distribution of can be described by its center, spread, and shape. For most problems, the central limit theorem allows us to conclude that the shape of the distribution is approximately normal, and we can use methods from chapter 4 to answer questions about the probability of the sample mean falling in various intervals.

An example at the end of the section demonstrates the type of exercise you should be able to solve.

#### The Mean

The mean of the sampling distribution of is the population mean . This is the balancing point of the sampling distribution.

#### The Standard Deviation

The standard deviation of the sampling distribution of , also know as the standard error of , or SE(), is smaller than the population standard deviation. This reflects the fact that the mean from a sample is more likely to be close to the population mean than a randomly chosen individual.

The formula for SE() is . The standard error may usually be thought of as a typical size for the distance between and due to chance.

#### The Shape

The shape of the sampling distribution of will be approximately normal (bell-shaped) for sufficiently large samples. For many sets of data arising in practice, "sufficiently large" might mean 25 or 30. Populations where "sufficiently large" needs to be substantially larger than 25 or 30 will generally be strongly skewed.

#### The Central Limit Theorem

The central limit theorem summarizes the information about the center, spread, and shape presented above.

For every population with a population mean and population standard deviation , even if the shape of the population is not normal, the sampling distribution of will be approximately normal with a mean of and a standard deviation of for large samples.

#### Example

In a recent year, the birthweights of infants born in Boston had a mean weight of 112.0 ounces with a standard deviation of 20.6 ounces. If 40 infants were randomly sampled, what is the probability that the mean of their weights would be between 105 and 115 ounces?

Draw a sketch!

By the central limit theorem, the sampling distribution of should be approximately normally distributed with a mean of = 112.0 and an SE of = 3.26. The two border points have z-scores of

z = (105 - 112)/3.26 = -2.15 and z = (115 - 112)/3.26 = 0.92.

The area between these points on the standard normal curve is

.8212 - .0158 = .8054, so there is about an 80% chance that the sample mean would be between 105 and 115.