SOLUTION: A certain brand of apple juice is supposed to have 64 oz. of juice. The juice varies from bottle to bottle though with a standard deviation of .06. The average amount of juice shou

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Question 1118673: A certain brand of apple juice is supposed to have 64 oz. of juice. The juice varies from bottle to bottle though with a standard deviation of .06. The average amount of juice should be 64 oz though. After randomly selecting a bottle off the line the manager says the content is about 1.5 standard deviations below average. Approximately how much juice is in the bottle?
As part of her routine, she now samples 22 bottles of juice. The results are as followed:
63.97, 63.92, 63.93, 63.87, 63.94, 63.98, 64.03, 63.9, 63.95, 64.05, 64.02, 63.9, 63.95, 64.01, 64.01, 63.91, 63.9, 63.92, 64, 63.93, 64.01, 63.97
The sampling distribution of the sample must be normal. Why can she be confident that the normality requirement has been met here?
The sample mean of the sample statistics above is 63.958. What is the s, and n of the sample statistics above?
What is the mu and sigma of the population parameters above?
Answer by stanbon(75887) About Me  (Show Source):
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A certain brand of apple juice is supposed to have 64 oz. of juice. The juice varies from bottle to bottle though with a standard deviation of .06. The average amount of juice should be 64 oz though. After randomly selecting a bottle off the line the manager says the content is about 1.5 standard deviations below average. Approximately how much juice is in the bottle?
Ans: 64-1.5*0.06 = 63.91 oz
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As part of her routine, she now samples 22 bottles of juice. The results are as followed:
63.97, 63.92, 63.93, 63.87, 63.94, 63.98, 64.03, 63.9, 63.95, 64.05, 64.02, 63.9, 63.95, 64.01, 64.01, 63.91, 63.9, 63.92, 64, 63.93, 64.01, 63.97
The sampling distribution of the sample must be normal. Why can she be confident that the normality requirement has been met here?
Ans: She selected a random sample.
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The sample mean of the sample statistics above is 63.958. What is the s, and n of the sample statistics above?
Ans: n = 22 ; Using my TI-84 I get::s = 7.0296
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What is the mu and sigma of the population parameters above?
Ans: mu = 63.958
Since sigma/sqrt(n) = 7.0296, sigma = sqrt(22)*7.0296 = 32.97
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Cheers,
Stan H.
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