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2010 Midterm Exam Solution V.2

# 2010 Midterm Exam Solution V.2 - Midterm.2 solution Q1...

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Midterm.2 solution February, 2010 MATH 203 Q1 Solution : (i) Roughly 50% since the 12 and 16 correspond roughly to the 25th and 75th percentiles. (ii) Roughly 12.5 (iii) Males seems to have higher variability than the females since it appears that there is greater variability between the 25th and 75th percentiles and the whiskers are larger. (iv) 3 (v) The observations seem to be skewed towards the left (lower values) and hence the mean is not a great measure of centrality. Q2 Solution : (i) Clearly a fish-in-lake problem, with N = 1000 , a = . 10 × 1000 = 100 and n = 50. P ( at least 3 contaminated ) = 1 - P ( less than 3 contaminated ) = 1 - [ P (0) + P (1) + P (2)] = 1 - ( 100 0 )( 900 50 ) ( 1000 50 ) + ( 100 1 )( 900 49 ) ( 1000 50 ) + ( 100 2 )( 900 48 ) ( 1000 50 ) (ii) Want P ( N 1 N 2 C 3 ) = P ( C 3 | N 1 N 2 ) P ( N 1 N 2 ) = P ( C 3 | N 1 N 2 ) P ( N 1 | N 2 ) P ( N 1 ) = 100 998 . 899 999 . 900 1000 = . 0812 - using the multiplication rule (twice) for conditional probability and conditioning backwards. Q3 Solution : Let EBV = event of exposure to EBV. MS = event of having MS. Given : P(EBV) = 0.95, P(MS) = .001, P ( EBV MS ) = . 9505 (i) P ( EBV MS ) = P ( EBV ) + P ( MS ) - P ( EBV MS ). (Addition rule) Therefore, P ( EBV

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