1. What if the Original Distribution Is Not Normal? Use the Central Limit Theorem.

2. Central Limit Theorem If x has any distribution with mean  and standard deviation , then the sample mean based on a random sample of size n will have a distribution that approaches the normal distribution (with mean  and standard deviation  divided by the square root of n) as n increases without bound.

3. How large should the sample size be to permit the application of the Central Limit Theorem? In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.

4. Central Limit Theorem

5. For most x distributions, if we use a sample size of 30 or larger, the distribution will be approximately normal.

6. Central Limit Theorem The mean of the sampling distribution is the same as the mean of the original distribution. The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

7. Central Limit Theorem Formula

10. Application of the Central Limit Theorem Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be: a.more than 530 pounds? b.less than 500 pounds? c.between 495 and 515 pounds?

11. Are we authorized to use the Normal Distribution? Yes, we are attempting to draw conclusions about means of large samples.

12. Applying the Central Limit Theorem What is the probability that their mean weight will be more than 530 pounds? Consider the distribution of sample means: P( x > 530): z = 530 – 510 = 20 = 2.22 9 9 P(z > 2.22) = _______.0132

13. Applying the Central Limit Theorem What is the probability that their mean weight will be less than 500 pounds? P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9 P(z < – 1.11) = _______.1335

14. Applying the Central Limit Theorem What is the probability that their mean weight will be between 495 and 515 pounds? P(495 < x < 515) : for 495: z = 495 – 510 =  15 =  1.67 9 9 for 515: z = 515 – 510 = 5 = 0.56 9 9 P(  1.67 < z < 0.56) = _______.6648