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stat100_midterm_spring2011_solutions

# stat100_midterm_spring2011_solutions - Statistics 100...

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Statistics 100 Midterm Spring 2011 9 March 2011 Solutions 1. Short answer questions . (a) A and B are not disjoint. Pr( B | A ) = Pr( B ) implies A and B are independent, so Pr( A and B ) = Pr( A ) Pr( B ) > 0 . If A and B were disjoint, Pr( A and B ) = 0 . (b) Solved by transforming to a standard Normal. Let W be the weight of a 12-month old girl. Pr(8 W 10) = Pr( W 10) - Pr( W 8) = Pr W - 9 2 10 - 9 2 - Pr W - 9 2 8 - 9 2 = Pr( Z 0 . 5) - Pr( Z ≤ - 0 . 5) = (1 - 0 . 3085) - 0 . 3085 = 0 . 38 (c) From the Normal tables, the lower 10 th percentile of a standard Normal is -1.28. The lower 10 th percentile of the weight distribution W is found by calculating w = μ + σz = 9 + 2 z = 9 + 2( - 1 . 28) = 6 . 44 kg (d) The IQR for the distribution is 2 . 575 - 0 . 78 = 1 . 795 . The distance to look beyond the first and third quartiles is 1 . 5(1 . 795) = 2 . 6925 . Q 1 - 2 . 6925 < 0 so there are no outliers below the first quartile. Q 3 + 2 . 6925 = 5 . 2625 so 9 . 51 and 11 . 39 are outliers. These correspond to 9 , 510 , 000 and 11 , 390 , 000 users. (e) Since the mean is much bigger than the median (1.01), the distribution is very likely right-skewed.

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