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Chapter 8: Problem 13

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5 . a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. b. What is the approximate probability that \(\bar{x}\) will be within \(0.5\) of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

Short Answer

Expert verified
a. The mean of the \(\bar{x}\) sampling distribution is 40, the standard deviation of the \(\bar{x}\) sampling distribution is 0.625, and the shape is approximately normal. b. The approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) is 0.5764. c. The approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than 0.7 is 0.484.

Step by step solution

01

Mean and Standard deviation of the \(\bar{x}\)

The mean of the sampling distribution (which is the mean of the means) is equal to the population mean \(\mu\), which in this case is 40. The standard deviation of the sampling distribution, also called the standard error \(SE\), is equal to the standard deviation of the population \(\sigma\) divided by the square root of the sample size \(n\), so: \(SE = \sigma / \sqrt{n} = 5 / \sqrt{64} = 5 / 8 = 0.625.\)
02

Describe the shape of the sampling distribution

The Central Limit Theorem tells us that the sampling distribution will be approximately normal when the sample size \(n\) is large, typically \(n > 30\). In this case, our sample size is 64, which is large, so we can say the sampling distribution will be approximately normal in shape.
03

Probabilities of \(\bar{x}\) being within a certain range

To find the probabilities, first convert the range to z-scores. A z-score tells us how many standard deviations away from the mean the value is. Use the formula: \(Z = (\bar{x} - \mu) / SE \). \nFor part b, the range is from 39.5 to 40.5. So calculate the z-scores: \(Z_{40.5} = (40.5 - 40) / 0.625 = 0.8\) and \(Z_{39.5} = (39.5 - 40) / 0.625 = -0.8\). The probability that \(\bar{x}\) is between these two values is the area under the standard normal curve between these two z-scores. This value is approximately 0.5764.\nFor part c, in a similar fashion, calculate the z-scores for range \(\mu - 0.7\) and \(\mu + 0.7\). Then, find the area under the curve, which represents the probability. The area is 1 minus the combined area for those two z-scores. That is, \(1 - 2*P(Z_{0.7})\). Since \(P(Z_{0.7})\) is 0.758, the result is approximately 0.484.

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