1. SAMPLING DISTRIBUTIONS
CHAPTER 7 (Part B)
SAMPLING DISTRIBUTIONS
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2. 7.3 SHAPE OF THE SAMPLING DISTRIBUTION OF x
The population from which samples are drawn has a normal distribution.
The population from which samples are drawn does not have a normal distribution.
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3. Sampling From a Normally Distributed Population
If the population from which the samples are drawn is normally distributed with mean μ and standard deviation σ, then the sampling distribution of the sample mean, , will also be normally distributed with the following mean and standard deviation, irrespective of the sample size:
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4. Figure : Population distribution and sampling distributions of .
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5. Example
In a recent SAT, the mean score for all examinees was Assume that the distribution of SAT scores of all examinees is normal with the mean of 1020 and a standard deviation of Let be the mean SAT score of a random sample of certain examinees. Calculate the mean and standard deviation of and describe the shape of its sampling distribution when the sample size is
(a) (b) (c) 1000
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6. Example 7-3: Solution (a) μ = 1020 and σ = 153.
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7. Figure
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8. Example: Solution
(b)
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9. Figure
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10. Example: Solution
(c)
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11. Figure
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12. Sampling From a Population That Is Not Normally Distributed
Central Limit Theorem
According to the central limit theorem, for a large sample size, the sampling distribution of is approximately normal, irrespective of the shape of the population distribution. The mean and standard deviation of the sampling distribution of are
The sample size is usually considered to be large if
n ≥ 30.
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13. Figure: Population distribution and sampling distributions of .
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14. Example
The mean rent paid by all tenants in a small city is $1550 with a standard deviation of $225. However, the population distribution of rents for all tenants in this city is skewed to the right. Calculate the mean and standard deviation of and describe the shape of its sampling distribution when the sample size is
(a) (b) 100
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15. Example: Solution
(a) Let x be the mean rent paid by a sample of 30 tenants.
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16. Figure
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17. Example: Solution
(b) Let x be the mean rent paid by a sample of 100 tenants.
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18. Figure
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19. 7.4 APPLICATIONS OF THE SAMPLING DISTRIBUTION OF x
If we take all possible samples of the same (large) size from a population and calculate the mean for each of these samples, then about 68.26% of the sample means will be within one standard deviation of the population mean.
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20. Figure
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21. APPLICATIONS OF THE SAMPLING DISTRIBUTION OF x
If we take all possible samples of the same (large) size from a population and calculate the mean for each of these samples, then about 95.44% of the sample means will be within two standard deviations of the population mean.
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22. Figure
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23. APPLICATIONS OF THE SAMPLING DISTRIBUTION OF x
If we take all possible samples of the same (large) size from a population and calculate the mean for each of these samples, then about 99.74% of the sample means will be within three standard deviations of the population mean.
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24. Figure
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25. Example
Assume that the weights of all packages of a certain brand of cookies are normally distributed with a mean of 32 ounces and a standard deviation of .3 ounce. Find the probability that the mean weight, , of a random sample of 20 packages of this brand of cookies will be between 31.8 and 31.9 ounces.
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26. Example: Solution
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27. z Value for a Value of x The z value for a value of is calculated as
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28. Example: Solution For = 31.8: For = 31.9:
P(31.8 < < 31.9) = P(-2.98 < z < -1.49)
= P(z < -1.49) - P(z < -2.98)
= = .0667
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29. Figure
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30. Example
According to Sallie Mae surveys and Credit Bureau data, college students carried an average of $3173 credit card debt in Suppose the probability distribution of the current credit card debts for all college students in the United States is known but its mean is $3173 and the standard deviation is $750. Let be the mean credit card debt of a random sample of 400 U.S. college students.
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31. Example
What is the probability that the mean of the current credit card debts for this sample is within $70 of the population mean?
What is the probability that the mean of the current credit card debts for this sample is lower than the population mean by $50 or more?
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32. Example: Solution
μ = $3173 and σ = $750. The shape of the probability distribution of the population is unknown. However, the sampling distribution of is approximately normal because the sample is large (n > 30).

33. Example: Solution
(a)
P($3103 ≤ ≤ $3243) = P(-1.87 ≤ z ≤ 1.87) = = .9386
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34. Figure
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35. Example: Solution
(a) Therefore, the probability that the mean of the current credit card debts for this sample is within $70 of the population mean is
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36. Example: Solution (b) For = $3123: P( ≤ 3123) = P (z ≤ -1.33) = .0918
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37. Figure
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