SOLUTION: 1. Given a normally distributed population with mean of 100 and a standard deviation of 20, find P(96 ≤ x-bar ≤ 108). Sample size is 16.
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Question 934751: 1. Given a normally distributed population with mean of 100 and a standard deviation of 20, find P(96 ≤ x-bar ≤ 108). Sample size is 16. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Given a normally distributed population with mean of 100 and a standard deviation of 20, find P(96 ≤ x-bar ≤ 108). Sample size is 16.
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z(96) = (96-100)/(20/sqrt(16)) = -4/5
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z(108) = (108-100)/(20/sqrt(16)) = 8/5
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P(96<= x-bar <=108) = P(-4/5<= z <=8/5) = 0.7334
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Cheers,
Stan H.
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