Question 552142
1.) Assume that the mean score on a certain aptitude test across the nation is 100, and that the standard deviation is 20 points. Find the probability that the mean aptitude test score for a randomly selected group of 150 8th graders is between 94 and 106.
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z(94) = (94-100)/20 = -6/20 = -0.3
z(106) = (106-100)/20 = 0.3
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P(94 < x < 106) = P(-0.3 < z < 0.3) = 0.2358
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2.) A 98% confidence interval estimate for a population mean was computed to be (33.7, 54.5). Determine the mean of the sample, which was used to determine the interval estimate (show all work).
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Solve:
xbar-ME = 33.7
xbar+ME = 54.5
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Add and solve for xbar:
2xbar = 88.2
x-bar = 44.1
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3.) A confidence interval estimate for the population mean is given 
to be (39.90 , 48.11). 
If the standard deviation is 11.473 and the sample size is 52, answer the following (show all work). 
(a) determine the maximum error of the estimate, E. 
Solve for "E"
xbar-E = 39.9
xbar+E = 48.11
Subtract and solve for "E":
2E = 8.21
E = 4.105
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(b) determine the confidence level used for the given confidence interval.
E = z*s/sqrt(n)
4.105 = z*11.473/sqrt(52)
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z = 4.105*sqrt(52)/11.473
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z = 2.5801
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P(z > 2.5801) = 0.0049
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Rounding you get Prob = 0.05
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Confidence level = 1-2*0.05 = 99%
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Cheers,
Stan H.
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