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Unformatted text preview: sampling distribution of xbar is nearly normal. (Or words to this effect, as long as the point is made that this is a large sample, larger than the n = 30 required by the CLT.) © 2008 Harvey A. Singer 1 b. Suppose a single random sample of n = 48 employees was selected. What is the chance that the mean of this sample will be within 2.5 years of the population mean? Want Prob ( μ2.5 years < xbar < μ +2.5 years) = Prob ( 41.2 years ≤ xbar ≤ 46.2 years) For xbar = 46.2 years, z = (xbar – μ )/ σ xbar = 2.5/1.198 = 2.09 So Prob ( 41.2 years ≤ xbar ≤ 46.2 years) = Prob (2.09 < z < 2.09) = Prob ( z ≤ 2.09) – Prob (z ≤ 2.09) = 0.9817 +0.0183 = 0.9634 Answer: 0.9634 c. What can the insurance carrier do to reduce the standard error? Focus your answer on issues related only to sampling concepts. To reduce standard error of the mean, increase the sample size: take samples of size greater than 48 employees. © 2008 Harvey A. Singer 2...
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 Fall '08
 SINGER
 Normal Distribution, Standard Deviation, Standard Error, Harvey A. Singer, Mean Sample Problem

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