Practice problems for these concepts can be found at:

- Binomial Distributions, Geometric Distributions, and Sampling Distributions Multiple Choice Practice Problems for AP Statistics
- Binomial Distributions, Geometric Distributions, and Sampling Distributions Free Response Practice Problems for AP Statistics
- Binomial Distributions, Geometric Distributions, and Sampling Distributions Cumulative Review Problems
- Binomial Distributions, Geometric Distributions, and Sampling Distributions Rapid Review for AP Statistics

Suppose we drew a sample of size 10 from a normal population with unknown mean and standard deviation and got = 18.87. Two questions arise: (1) what does this sample tell us about the population from which the sample was drawn, and (2) what would happen if we drew more samples?

Suppose we drew 5 more samples of size 10 from this population and got = 20.35, = 20.04, = 19.20, = 19.02, and = 20.35. In answer to question (1), we might believe that the population from which these samples was drawn had a mean around 20 because these averages tend to group there (in fact, the six samples were drawn from a normal population whose mean is 20 and whose standard deviation is 4). The mean of the 6 samples is 19.64, which supports our feeling that the mean of the original population might have been 20.

The standard deviation of the 6 samples is 0.68 and you might not have any intuitive sense about how that relates to the population standard deviation, although you might suspect that the standard deviation of the samples should be less than the standard deviation of the population because the chance of an extreme value for an average should be less than that for an individual term (it just doesn't seem very likely that we would draw a *lot* of extreme values in a single sample).

Suppose we continued to draw samples of size 10 from this population until we were exhausted or until we had drawn *all possible samples of size 10*. If we did succeed in drawing all possible samples of size 10, and computed the mean of each sample, the distribution of these sample means would be the **sampling distribution of** .

Remembering that a "statistic" is a value that describes a sample, the **sampling distribution of a statistic** is the distribution of that statistic for all possible samples of a given size. It's important to understand that a dotplot of a few samples drawn from a population is not a distribution (it's a *simulation* of a distribution)—it becomes a distribution only when all possible samples of a given size are drawn.

### Sampling Distribution of a Sample Mean

Suppose we have the sampling distribution of . That is, we have formed a distribution of the means of all possible samples of size *n* from an unknown population (that is, we know little about its shape, center, or spread). Let μ_{} and σ_{} represent the mean and standard deviation of the sampling distribution of , respectively.

Then

for any population with mean μ and standard deviation σ.

(Note: the value given for above is generally considered correct only if the sample size (*n*) is small relative to *N*, the number in the population. A general rule is that *n* should be no more than 5% of *N* to use the value given for σ_{} (that is, *N* > 20*n*). If *n* is more than 5% of *N*, the exact value for the standard deviation of the sampling distribution is

In practice this usually isn't a major issue because

is close to one whenever *N* is large in comparison to *n*. You don't have to know this for the AP exam.)

example:A large population is know to have a mean of 23 and a standard deviation of 2.5. What are the mean and standard deviation of the sampling distribution of means of samples of size 20 drawn from this population?

solution:

### Central Limit Theorem

The discussion above gives us measures of center and spread for the sampling distribution of but tells us nothing about the *shape* of the sampling distribution. It turns out that the shape of the sampling distribution is determined by (a) the shape of the original population and (b) *n*, the sample size. If the original population is normal, then it's easy: the shape of the sampling distribution will be normal if the population is normal.

If the shape of the original population is not normal, or unknown, and the sample size is small, then the shape of the sampling distribution will be similar to that of the original population. For example, if a population is skewed to the right, we would expect the sampling distribution of the mean for small samples also to be somewhat skewed to the right, although not as much as the original population.

When the sample size is large, we have the following result, known as the **central limit theorem:** For large *n*, the sampling distribution of will be approximately normal. The larger is *n*, the more normal will be the shape of the sampling distribution.

A rough rule-of-thumb for using the central limit theorem is that *n* should be at least 30, although the sampling distribution may be approximately normal for much smaller values of *n* if the population doesn't depart markedly from normal. The central limit theorem allows us to use normal calculations to do problems involving sampling distributions without having to have knowledge of the original population. Note that calculations involving *z*-procedures require that you know the value of σ, the population standard deviation. Since you will rarely know σ, the large sample size essentially says that the sampling distribution is approximately, but not exactly, normal. That is, technically you should not be using *z*-procedures unless you know σ but, as a practical matter, *z*-procedures are numerically close to correct for large *n*. Given that the population size (*N*) is large in relation to the sample size (*n*), the information presented in this section can be summarized in the following table:

example:Describe the sampling distribution of for samples of size 15 drawn from a normal population with mean 65 and standard deviation 9.

solution:Because the original population is normal, is normal with mean 65 and standard deviation = 2.32. That is, hasN.

example:Describe the sampling distribution of for samples of size 15 drawn from a population that is strongly skewed to the left (like the scores on a very easy test) with mean 65 and standard deviation 9.

solution:μ_{}and σ_{}= 2.32 as in the above example. However this time the population is skewed to the left. The sample size is reasonably large, but not large enough to argue, based on our rule-of-thumb (n ≥ 30), that the sampling distribution is normal. The best we can say is that the sampling distribution is probably more mound shaped than the original but might still be somewhat skewed to the left.

example:The average adult has completed an average of 11.25 years of education with a standard deviation of 1.75 years. A random sample of 90 adults is obtained. What is the probability that the sample will have a mean

- greater than 11.5 years?
- between 11 and 11.5 years?

solution:The sampling distribution of has μ_{}= 11.25 andBecause the sample size is large (

n= 90), the central limit theorem tells us that large sample techniques are appropriate. Accordingly,

- The graph of the sampling distribution is shown below:
- From part (a), the area to the left of 11.5 is 1 – 0.0869 = 0.9131. Since the sampling distribution is approximately normal, it is symmetric. Since 11 is the same distance to the left of the mean as 11.5 is to the right, we know that P( < 11) = P( > 11.5) = 0.0869. Hence, P(11 < <11.5) = 0.9131 – 0.0869 = 0.8262. The calculator solution is normalcdf(11,11.5,11, 0.184)=0.8258.

example:Over the years, the scores on the final exam for AP Calculus have been normally distributed with a mean of 82 and a standard deviation of 6. The instructor thought that this year's class was quite dull and, in fact, they only averaged 79 on their final. Assuming that this class is a random sample of 32 students from AP Calculus, what is the probability that the average score on the final for this class is no more than 79? Do you think the instructor was right?

solution:If this group really were typical, there is less than a 1% probability of getting an average this low by chance alone. That seems unlikely, so we have good evidence that the instructor was correct.

(The calculator solution for this problem is normalcdf(-1000,79, 82,1.06).)

Practice problems for these concepts can be found at:

- Binomial Distributions, Geometric Distributions, and Sampling Distributions Multiple Choice Practice Problems for AP Statistics
- Binomial Distributions, Geometric Distributions, and Sampling Distributions Free Response Practice Problems for AP Statistics
- Binomial Distributions, Geometric Distributions, and Sampling Distributions Cumulative Review Problems
- Binomial Distributions, Geometric Distributions, and Sampling Distributions Rapid Review for AP Statistics

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