Practice problems for these concepts can be found at:

- Sampling Distributions Solved Problems for Beginning Statistics
- Sampling Distributions Supplementary Problems for Beginning Statistics

If samples are selected from a population which is normally distributed with mean μ and standard deviation σ, then the distribution of sample means is normally distributed and the mean of this distribution is = μ, and the standard deviation of this distribution is . The shape of the distribution of the sample means is normal or bell-shaped regardless of the sample size.

The *central limit theorem* states that when sampling from a large population of any distribution shape, the sample means have a normal distribution whenever the sample size is 30 or more. Furthermore, the mean of the distribution of sample means is , and the standard deviation of this distribution is . It is important to note that when sampling from a nonnormal distribution, has a normal distribution only if the sample size is 30 or more. The central limit theorem is illustrated graphically in Figure 7-1.

Figure 7-1 illustrates that for samples greater than or equal to 30, has a distribution that is bell-shaped and centers at μ. The spread of the curve is determined by .

**EXAMPLE 7.12** If a large number of samples each of size n, where n is 30 or more, are selected and the means of the samples are calculated, then a histogram of the means will be bell-shaped regardless of the shape of the population distribution from which the samples are selected. However, if the sample size is less than 30, the histogram of the sample means may not be bell-shaped unless the samples are selected from a bell-shaped distribution.

### Applications of the Sampling Distribution of the Sample Mean

The distribution properties of the sample mean are used to evaluate the results of sampling, and form the underpinnings of many of the statistical inference techniques found in the remaining chapters of this text. The examples in this section illustrate the usefulness of the central limit theorem.

**EXAMPLE 7.13** A government report states that the mean amount spent per capita for police protection for cities exceeding 150,000 in population is $500 and the standard deviation is $75. A criminal justice research study found that for 40 such randomly selected cities, the average amount spent per capita for this sample for police protection is $465. If the government report is correct, the probability of finding a sample mean that is $35 or more below the national average is given by P( < 465). The central limit theorem assures us that because the sample size exceeds 30, the sample mean has a normal distribution. Furthermore, if the government report is correct, the mean of is $500 and the standard error is = $11.86. The area to the left of $465 under the normal curve for is the same as the area to the left of Z = . We have the following equality: P( < 465) = P(Z < –2.95) = .0016. This result suggests that either we have a highly unusual sample or the government claim is incorrect. Figures 7-2 and 7-3 illustrate the solution graphically.

In Example 7.13, the value 465 was transformed to a z value by subtracting the population mean 500 from 465, and then dividing by the standard error 11.86. The equation for transforming a sample mean to a z value is shown in formula (*7.6*):

The solution to Example 7.13 using EXCEL would look like the following:

The first three rows are EXCEL messages reiterating the Central Limit Theorem for the sample mean. The fourth row contains a message, a computation, and a message explaining the computation. The function, =NORMDIST(465,500,11.86,1), asks EXCEL to find the area under the normal curve to the left of 465. The curve is centered at 500 and has standard deviation equal to 11.86. The 1 in the fourth position tells EXCEL to find the cumulative area or the area to the left of 465. See Figure 7-2.

**EXAMPLE 7.14** A machine fills containers of coffee labeled as 113 grams. Because of machine variability, the amount per container is normally distributed with μ = 113 and σ = 1. Each day, 4 of the containers are selected randomly and the mean amount in the 4 containers is determined. If the mean amount in the four containers is either less than 112 grams or greater than 114 grams, the machine is stopped and adjusted. Since the distribution of fills is normally distributed, the sample mean is normally distributed even for a sample as small as four. The mean of the sample mean is = 113 and the standard error is = .5. The machine is adjusted if < 112 or if > 114. The probability that the machine is adjusted is equal to the sum P( < 112) + P( > 114) since we add probabilities of events connected by the word *or*. To evaluate these probabilities, we use formula (*7.6*) to express the events involving in terms of z as follows:

The probability that the machine is adjusted is 2 × .0228 = .0456. It is seen that if this sampling technique is used to monitor this process, there is a 4.56% chance that the machine will be adjusted even though it is maintaining an average fill equal to 113 grams.

Practice problems for these concepts can be found at:

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