Question 456928
A population has a normal distribution with a mean of 130 and a standard deviation of 30. Find the probability that a single element selected from the population will have a value between 139.50 and 167.25.
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z(139.5)=(139.5-130)/30 = 0.3167
z(167.25)=(167.25-130)/30 = 1.2417
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P(139.5< x < 167.25) = P(0.3167< z <1.2417) = 0.2686
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Use the population described in problem 2 above. Now find the probability that the sample mean for a sample of 16 elements selected from the population will be between 139.50 and 167.25.
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t(139.5)=(139.5-130)/[30/sqrt(16)] = 1.2667
t(167.25)=(167.25-130)/[30/sqrt(16)] = 4.9667
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P(139.5 < x-bar < 167.25) = P(1.2667 < t < 4.9667 when df = 15) = 0.1122
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Cheers,
Stan H.
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