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Unformatted text preview: APPM 4/5570 Solutions to Problems from Section 4.3 35. X N (8 . 8 , 2 . 8 2 ), ie: = 8 . 8, 2 = 2 . 8 2 , and = 2 . 8. a. P ( X 10) = P X 8 . 8 2 . 8 10 8 . 8 2 . 8 P ( Z . 43) = 1 (0 . 43) = ( . 43) = 0 . 3336 P ( X > 10) = P ( X 10) = 0 . 3336 b. P ( X > 20) = P X 8 . 8 2 . 8 > 20 8 . 8 2 . 8 P ( Z > 4) 0 given the accuracy of Table A. 3. c. P (5 X 10) = P 5 8 . 8 2 . 8 X 8 . 8 2 . 8 10 8 . 8 2 . 8 P ( 1 . 36 Z . 43) = (0 . 43) ( 1 . 36) = 0 . 6664 . 0869 = 0 . 5795 d. Want to solve P (8 . 8 c < X < 8 . 8 + c ) = 0 . 98 for c . Standardizing, this gives P 8 . 8 c 8 . 8 2 . 8 < X 8 . 8 2 . 8 < 8 . 8+ c 8 . 8 2 . 8 = P c 2 . 8 < Z < c 2 . 8 . By symmetry about zero of the zcurve, we want to find c so that P Z < c 2 . 8 = 0 . 98 + 0 . 01 = 0 . 99 . A reverse lookup in Table A. 3 gives us that c 2 . 8 2 . 33 ....
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This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.
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