Stat 5131, Fall 1997 (Geyer) Second Midterm Solutions

# Stat 5131, Fall 1997 (Geyer) Second Midterm Solutions - f Y...

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Up: Stat 5101 Stat 5131 Midterm 2 N = 24 Median = 68.5 Quartiles = 47, 87 grad undergrad 3 : 14 3 : 14 3 : 5 3 : 5 4 : 2 4 : 2 4 : 577 4 : 577 5 : 4 5 : 4 5 : 5 : 6 : 223 6 : 3 6 : 22 6 : 6 : 6 : 7 : 1144 7 : 1 7 : 144 7 : 56 7 : 5 7 : 6 8 : 2 8 : 8 : 2 8 : 8 : 8 : 9 : 24 9 : 4 9 : 2 9 : 57 9 : 57 9 : 10 : 00 10 : 0 10 : 0 Problem 1 Median = 20, Lower Quartile = 19, Upper Quartile = 20 (a) The distribution of X is symmetric about zero, hence E ( X ) = 0by Theorem 4 of Chapter 4 in Lindgren. (b) Since E ( X ) = 0 by part (a),

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Problem 2 Median = 20, Lower Quartile = 16, Upper Quartile = 20 The conditional p. d. f. of X given Y is So to find f ( x | y ) we first have to find the marginal

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Unformatted text preview: f Y ( y ). (Note: this was done in problem 4 of the first midterm. This part of the question is a rehash.) To find the marginal of Y , integrate out X Then Problem 3 Median = 0, Lower Quartile = 0, Upper Quartile = 16.5 The regression function of Y on X is another name for E ( Y | X ). From the formula given in the hint so the regression function is Problem 4 Median = 19, Lower Quartile = 9, Upper Quartile = 20 (a) (b) Problem 5 Median = 11.5, Lower Quartile = 4, Upper Quartile = 16 Evaluating at zero we get If we write , then and Up: Stat 5101 Charles Geyer 1999-10-25...
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## Stat 5131, Fall 1997 (Geyer) Second Midterm Solutions - f Y...

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