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Asst8-suggested - Statistics 427: Homework #! Solutions to...

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Statistics 427: Homework # ! Solutions to Suggested Problems Note : This page contains solutions for 4.59 and 4.67; the following pages contain solutions for 4.65, 4.68, 5.3 and 5.13. 4.59 X is Gamma with ! = 1 and " = 1 a. E(X) = 1 1 = ! b. 1 1 = = " c. P(X ! 4 ) = 982 . 1 1 4 ) 4 )( 1 ( = ! = ! ! ! e e d. P(2 ! X ! 5) = [ ] 129 . 1 1 5 2 ) 2 )( 1 ( ) 5 )( 1 ( = ! = ! ! ! ! ! ! ! e e e e 4.67 Suppose μ = 24 and " 2 = 144 # $% = 24, $% 2 = 144 # % = 6, $ = 4 a. P(12 ! X ! 24 ) = F(4;4) – F(2;4) = .424 b . P(X ! 24) = F(4;4) = .567, so while the mean is 24, the median is less than 24. (Note that P(X ! μ ~ ) = .5) The fact that the median is less than the mean is caused by the positive skew of the gamma distribution. c . We want a value of X for which F(X; 4) =.99. From Table A.4, we see F(10;4)=.990. So with % = 6, the 99 th percentile = 6(10)=60. d . We want a value of X for which F(X; 4) =.995. Based on Table A.4, F(11; 4) = .995, so t = 6(11)=66. Thus at 66 weeks, only .5% of all transistors would still be operating.
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4.65 We are given that X Gamma( α =7 , β = 1). (a) P ( X 5) = F (5; α = 7) = 0 . 238 (from table) (b) P ( X< 5) = P ( X 5) = 0 . 238 (c) P ( X> 8) = 1 - P ( X 8) = 1 - F (8; α = 7) = 1 - 0 . 687 = 0 . 313 (d) P (3 X 8) = P ( X 8) - P ( 3) = F (8; α = 7) - F (3; α = 7) = 0 . 687 - 0 . 034 = 0 . 653 (e) P (3 < X < 8) = P (3 X 8) = 0 . 653 (f) P ( 4 or 6) = P
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This note was uploaded on 04/07/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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Asst8-suggested - Statistics 427: Homework #! Solutions to...

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