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Unformatted text preview: 7. a. E(Y) = . 6 . 5 7 . 8 = × = × π n b. Under the assumption that 70% of the customers are satisfied, the onetailed probability to observe 2 or less successes is ) 2 ( ) 1 ( ) ( P P P + + = 6 2 7 1 8 3 . 7 . 2 8 3 . 6 . 1 8 3 . 7 . 8 + + . The result is 01129 . 0.01000 0.00122 0.00007 ( = + + .) Since this probability is smaller than 0.05, there is strong reason to doubt the restaurant’s claim. (Note that using empirical rule for this problem is not that good since when number of failures is less than 5, Normal approximation does not work here .) One can also use minitab to figure out the probabilities. 8. mean of sample mean = 3, standard deviation of sample mean = 1/sqrt(9) = 1/3 P(sample mean >3.1) = P( (Ybar – 3)/.333) > (3.13/.333) ) =P(Z>.3) =1.6179=.3821 or 38.21%...
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 Spring '08
 Chow
 Statistics, Normal Distribution, Standard Deviation, 40%, 38.21%

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