Unformatted text preview: What should be the approximate mean and standard deviation of the distribution of the mean amount of beer of these 16 students? Mean will still be 16, i.e. X =16, also known as expected amount beer is 16 SE( X ) = 16 16 n =16/4=4 b. For these 16 students, there is about a 68% chance that the mean amount of beer will be between ___ and ___. 16 16 * 1 40 =(404, 40+4)=(36, 44). c. If the sample mean these 16 students was 48 ouches, use the Standard Normal Table to find P( X >48). P( X > 48) = P ) 00 . 2 ( 16 16 40 48 Z P Z P n X Z = 1P(Z<2.00)=10.9772=0.0228 d. Write an interpretation for this outcome in part ii. There is approximately a 2.3% chance that a sample of 16 students would have a mean of more than 48 ouches....
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 Fall '08
 BARROSO,JOAOR
 Statistics, Normal Distribution, Standard Deviation, sample proportion, State College police, PSU students, normal approximation methods

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