<PRE><B>Question 1118673</B>
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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