Question 279663: Procter & Gamble reported that an American family of four washes an average of 1 ton (2000 lbs) of clothes each year. If the standard deviation is 187.5 lb, find the probability that the mean of a randomly selected sample of 50 families of four will be between 1980 and 1990 lbs.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Procter & Gamble reported that an American family of four washes an average of 1 ton (2000 lbs) of clothes each year. If the standard deviation is 187.5 lb, find the probability that the mean of a randomly selected sample of 50 families of four will be between 1980 and 1990 lbs.
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t(1990) = (1990-2000)/[187.5/sqrt(50)] = -0.3771
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t(1989) = (1980-2000)/[187.5/sqrt(50)] = -0.7542
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P(1980< x < 1990) = P(-0.7542< t < -0.3771)
= tcdf(-0.7542,-0.2771,49) = 0.1267
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Cheers,
Stan H.
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