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M ARIO F . T RIOLA

S TATISTICS. E LEMENTARY. Section 5-5 The Central Limit Theorem. M ARIO F . T RIOLA. E IGHTH. E DITION. Definition. Sampling Distribution of the mean is a probability distribution of sample means, with all samples having the same sample size n.

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M ARIO F . T RIOLA

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  1. STATISTICS ELEMENTARY Section 5-5 The Central Limit Theorem MARIO F. TRIOLA EIGHTH EDITION

  2. Definition Sampling Distribution of the mean is a probability distribution of sample means, with all samples having the same sample size n.

  3. 1. The random variable xhas a distribution (which may or may not be normal) with mean µ and standard deviation . 2. Samples all of the same size n are randomly selected from the population of xvalues. Central Limit Theorem Given:

  4. Central Limit Theorem Conclusions: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means µx will be the population mean µ. 3. The standard deviation of the sample means xwill approach  n

  5. 1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample sizen (not just the values of n larger than 30). Practical Rules Commonly Used:

  6. the mean of the sample means the standard deviation of sample mean  (often called standard error of the mean) Notation µx= µ  x= n

  7. 0 Distribution of 200 digits from Social Security Numbers (Last 4 digits from 50 students) 20 Frequency 10 0 1 2 3 4 5 6 7 8 9 Distribution of 200 digits Figure 5-19

  8. 1 5 9 5 9 4 7 9 5 7 8 3 8 1 3 2 7 1 3 6 3 8 2 3 6 1 5 3 4 6 4 6 8 5 5 2 6 4 9 4.75 4.25 8.25 3.25 5.00 3.50 5.25 4.75 5.00 2 6 2 2 5 0 2 7 8 5 3 7 7 3 4 4 4 5 1 3 6 7 3 7 3 3 8 3 7 6 2 6 1 9 5 7 8 6 4 0 7 4.00 5.25 4.25 4.50 4.75 3.75 5.25 3.75 4.50 6.00 Table 5-2 SSN digits x

  9. 0 Distribution of 50 Sample Means for 50 Students 15 Frequency 10 5 0 1 2 3 4 5 6 7 8 9 Figure 5-20

  10. As the sample size increases, the sampling distribution of sample means approaches a normal distribution.

  11. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb.b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb.

  12. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. z = 150-143 = 0.24 29 P(x>150) = 0.405 Normalcdf(0.24,9999) = 0.4052 = 29 0.0948  = 143 150 0 0.24

  13. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb.

  14. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. P(x>150) =0.0735 z = 150-143 = 1.45 29 36 Normalcdf(1.45, 9999) = 0.0735 x= 4.83333 0.4265 x = 143 150 0 1.45

  15. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,

  16. a.) if one woman is randomly selected, find the probability    that her weight is greater than 150 lb.P(x > 150) = 0.4052 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,

  17. a.) if one woman is randomly selected, find the probability    that her weight is greater than 150 lb.P(x > 150) = 0.4052 b.) if 36 different women are randomly selected, their mean    weight is greater than 150 lb.P(x> 150) = 0.0735 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,

  18. a.) if one woman is randomly selected, find the probability    that her weight is greater than 150 lb.P(x > 150) = 0.4052 b.) if 36 different women are randomly selected, their mean    weight is greater than 150 lb.P(x > 150) = 0.0735It is much easier for an individual to deviate from the mean than it is for a group of 36 to deviate from the mean. Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb,

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