SOLUTION: Please help!
Of the total population of older Americans, 18% live in Florida. For a randomly selected sample of 200 older Americans, find the probability that more than 40 live
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Of the total population of older Americans, 18% live in Florida. For a randomly selected sample of 200 older Americans, find the probability that more than 40 live
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Question 168703: Please help!
Of the total population of older Americans, 18% live in Florida. For a randomly selected sample of 200 older Americans, find the probability that more than 40 live in Florida Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Of the total population of older Americans, 18% live in Florida. For a randomly selected sample of 200 older Americans, find the probability that more than 40 live in Florida
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The mean of the distribution of sample proportions is "np" = 0.18*200 = 36
The standard deviation of sqrt(npq) = sqrt(36*0.82) = 5.4332
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z(40) = (40-36)/5.4332 = 0.7362
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P(x>40) = P(z>0.7362) = 0.23000
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Cheers,
Stan H.
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