Question 195602
#1 problem:
The Old Farmer's Almanac reports that the average person uses 123 gallons of water daily. If the standard deviation is 21 gallons, find the probability that the mean of a randomly selected sample of 15 people will be between 120 and 126 gallons. Assume the variable is normally distributed. 
#2 problem:
Procter & Gamble reported that an American family of four washes an average of 1 ton (2000 pounds) of clothes each year. If the standard deviation of the distribution is 187.5 pounds, find the probability that the mean of a randomly selected sample of 50 families of four will be between 1980 and 1990 pounds. 
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Find the z-values of 1980 and 1990
z(1980) = (1980-2000)/[187.5/sqrt(50)]  = -0.7542
z(1990) = (1990-2000)/[187.5/sqrt(50)]  = -0.37712
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The P(1980 < xbar < 1990) = P(-0.7542 < z < -0.3771) = 0.1277
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#3 problem:
The average time it takes a group of adults to complete a certain achievement test is 46.2 minutes. The standard deviation is 8 minutes. Assume the variable is normally distributed.
a.) Find the probability that a randomly selected adult will complete the test in less than 43 minutes.
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z(43) = (43-46.2)/8 = -0.4
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P(x < 43) = P(z < -0.4) = 0.3446
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b.) Find the probability that, if 50 randomly selected adults take the test, the mean time it takes the group to complete the test will be less than 43 minutes.
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z(43) = (43-46.2)/[8/sqrt(50)] = -2.8284
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P(xbar < 43) = P(z < -2.8284) = 0.0023
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c.)Does it seem reasonable that an adult would finish the test in less than 43 minutes? Explain.
He has better than a 34% chance of doing that, as seen above.
Cheers,
Stan H.