Question

I need a brief summary of these things:

The Distribution of the Sample Mean

• The Case of a Normal Population Distribution
• The Central Limit Theorem

Answer 1

The Case of a Normal Population Distribution

Sampling distributions are an important part of study for a variety of reasons.
In most cases, the feasibility of an experiment dictates the sample size.
Sampling distribution is the probability distribution of a sample of a population
instead of the entire population.

In simpler words, suppose from a given population you take all possible samples
of size n and compute a statistic (say mean) of all these samples. If you then
prepare a probability distribution of this statistic, you will get a sampling distribution.

The properties of sampling distribution can vary depending on how small the sample
is as compared to the population. The population is assumed to be normally distributed as
is generally the case. If the sample size is large enough, the sampling distribution will
also be nearly normal.

If this is the case, then the sampling distribution can be totally determined by two
values - the mean and the standard deviation. These two parameters are important to compute
for the sampling distribution if we are given the normal distribution of the entire population.

Sampling Distribution of the Mean and Standard Deviation
Sampling distribution of the mean is obtained by taking the statistic under study of the
sample to be the mean. The say to compute this is to take all possible samples of sizes n
from the population of size N and then plot the probability distribution. It can be shown
that the mean of the sampling distribution is in fact the mean of the population.

The standard deviation however is different for the sampling distribution as compared to
the population. If the population is large enough, this is given by:

The Central Limit Theorem and Means

An essential component of the Central Limit Theorem is that the average of your sample means will be the population mean. In other words, add up the means from all of your samples, find the average and that average will be your actual population mean. Similarly, if you find the average of all of the standard deviations in your sample, you’ll find the actual standard deviation for your population. It’s a pretty useful phenomenon that can help accurately predict characteristics of a population. Watch a video explaining this phenomenon, or read more about it here: The Mean of the Sampling Distribution of the Mean.

General Steps
Step 1: Identify the parts of the problem. Your question should state:

1. the mean (average or μ)
2. the standard deviation (σ)
3. population size
4. sample size (n)
5. a number associated with “greater than” ( ). Note: this is the sample mean. In other words, the problem is asking you “What is the probability that a sample mean of x items will be greater than this number?

Step 2: Draw a graph. Label the center with the mean. Shade the area roughly above (i.e. the “greater than” area). This step is optional, but it may help you see what you are looking for.

Step 3: Use the following formula to find the z-score. Plug in the numbers from step 1.

Click here if you want easy, step-by-step instructions for solving this formula.

1. Subtract the mean (μ in step 1) from the ‘greater than’ value (in step 1). Set this number aside for a moment.
2. Divide the standard deviation (σ in step 1) by the square root of your sample (n in step 1). For example, if thirty six children are in your sample and your standard deviation is 3, then 3 / √36 = 0.5
3. Divide your result from step 1 by your result from step 2 (i.e. step 1/step 2)

Step 4: Look up the z-score you calculated in step 3 in the z-table. If you don’t remember how to look up z-scores, you can find an explanation in step 1 of this article: Area to the right of a z-score.

Step 5: Subtract your z-score from 0.5. For example, if your score is 0.1554, then 0.5 – 0.1554 = 0.3446.

Step 6: Convert the decimal in Step 5 to a percentage. In our example, 0.3446 = 34.46%.