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THE CENTRAL LIMIT THEOREM. Sampling Distribution of Sample Means Definition: A distribution obtained by using the means computed from random samples of.

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Presentation on theme: "THE CENTRAL LIMIT THEOREM. Sampling Distribution of Sample Means Definition: A distribution obtained by using the means computed from random samples of."— Presentation transcript:

1 THE CENTRAL LIMIT THEOREM

2 Sampling Distribution of Sample Means Definition: A distribution obtained by using the means computed from random samples of a specific size taken from a population. Project Time!

3 Properties of the Distribution of Sample Means 1.The mean of the sample means will be the same as the population mean 2.The standard deviation of the sample means will be smaller than the standard deviation of the population, in fact it will be equal to the population standard deviation divided by the square root of the sample size.

4 Standard Error of the Mean In addition to being the standard Deviation of the sampling distribution of sample means, it is also called The standard error of the mean Why do you think it is always smaller than the population standard deviation?

5 The Central Limit Theorem As the sample size n increases without limit, the shape of the distribution of the sample means taken with replacement from a population with mean  and standard deviation  will approach a normal distribution. As previously shown, this distribution will have a mean  and standard deviation of

6 Using the Central Limit Theorem Can be used in accordance with the normal distribution to answer questions about sample means. Modified z-score formula: or

7 Example 1: The average age of a vehicle registered in the U.S. is 8 years. Assume the standard deviation is 16 months. If a random sample of 36 cars is selected, find the probability that the mean of their age is between 90 and 100 months.

8 Solution Example 1: 1.Find z-scores for 90 and 100 months. z = -2.25z = 1.5

9 Solution Example 1 continued 2.Find the area under the normal curve between the 2 z-scores. @ z = -2.25 A =.4878 @ z = 1.5 A =.4332 Final Answer: A =.9210

10 Example 2 The average number of pounds of meat a person consumes a year is 218.4 pounds. Assume that the standard deviation is 25 pounds and the distribution is normal. a) Find the probability that a person selected at random consumes less than 224 pounds per year b) If a sample of 40 individuals is selected, find the probability that the mean of the sample will be less that 224 pounds per year.

11 Solution Example 2a: a)Find z-score for 224 lbs Z =.22 @ z =.22 A =.0871.5 +.0871 =.5871

12 Solution Example 2b: a)Find z-score for 224 lbs Z = 1.42 @ z = 1.42 A =.4222.5 +.4222 =.9222

13 Example 3 If the average sodium content of 150 microwave dinners is 795 mg, find the probability that a sample of 10 dinners will have an average sodium content less than 725mg. The standard deviation of population is 75mg.

14 Solution Example 3: a)Find z-score for 725 mg Z = -2.95 @ z = -2.95 A =.4984.5 -.4984 =.0016


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