solPracmid1 - 7. a. E(Y) = . 6 . 5 7 . 8 = = n b. Under the...

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Stat 500 : Solutions to Practice Exam 1 1. T 2. T 3. h 4. (1) mean of data set 1 is 7.22, median is 7.2 mean of data set 2 is 17.8, median is 7.8 (2) median is affected little by extreme measurements, whereas mean is influenced by extreme measurements (3) sample range of data set 1 is 4.0 sample range of data set 2 is 65.7 (4) s R/4 = 1 (5) Only 40% of the data fall within the interval (6.22, 8.22). Explanation: either the data set is too small to apply the empirical rule or R/4 is not a good estimate for s if the data set is too small. 5. P(M) = 0.4, P(W) = 0.5, P(M|W) = 0.7 (1) P(M W) = P(W) P(M|W) = 0.35 (2) P(W|M) = P(W M) = 0.35 = 0.875 P(M) 0.4 (3) P(W M ) + P(M W ) = P(W) – P(M W) + P(M) – P(M W) = 0.2 (4) P(W M) = 0.35 P(M) P(W) W, M not independent 6. y = number of persons cured out of 20 when given new medicine n = 20, π = 0.85, y has a binomial distribution (a) P(y = 20) = 20! (0.85) 20 (0.15) 0 = 0.85 20 = 0.03876 20! 0! (b) P(y = 0) = 20! (0.85) 0 (0.15) 20 = 0.15 20 = 3.3 × 10 -17 0! 20! Note: one can also use minitab to figure out these probabilities
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Unformatted text preview: 7. a. E(Y) = . 6 . 5 7 . 8 = = n b. Under the assumption that 70% of the customers are satisfied, the one-tailed 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 restaurants 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) > (3.1-3/.333) ) =P(Z>.3) =1-.6179=.3821 or 38.21%...
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solPracmid1 - 7. a. E(Y) = . 6 . 5 7 . 8 = = n b. Under the...

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