Question 195602: plS help me!.. URgent..I need your cooperatioN!
#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.
#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.
b.) Find the probability that, if 50 randomly selected adult take the test, the mean time it takes the group to complete the test will be less than 43 minutes.
c.)Does it seem reasonable that an adult would finish the test in less than 43 minutes? Explain.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! #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.
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