HW 1 Solutions

# HW 1 Solutions - HW 1 Solutions 1.11 a b c c = c c c = nc...

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HW 1 Solutions 1.11 a. = n i c 1 = c + c + … + c = nc . b. i n i y c = 1 = c ( y 1 + … + y n ) = = n i i y c 1 c. ( 29 = + n i i i y x 1 = x 1 + y 1 + x 2 + y 2 + … + x n + y n = ( x 1 + x 2 + … + x n ) + ( y 1 + y 2 + … + y n ) Using the above, the numerator of s 2 is = - n i i y y 1 2 ) ( = = + - n i i i y y y y 1 2 2 ) 2 ( = = - n i i y 1 2 = + n i i y n y y 1 2 2 Since = = n i i y y n 1 , we have = - n i i y y 1 2 ) ( = = - n i i y n y 1 2 2 . Let = = n i i y n y 1 1 to get the result. 2.16 a. 1/3 b. 1/3 + 1/15 = 6/15 c. 1/3 + 1/16 = 19/48 d. 49/240 2.76 Define the events: U : job is unsatisfactory A : plumber A does the job a. P ( U | A ) = P ( A U )/ P ( A ) = P ( A | U ) P ( U )/ P ( A ) = .5*.1/.4 = 0.125 b. From part a. above, 1 – P ( U | A ) = 0.875. 2.90 a. (1/50)(1/50) = 0.0004. b. P (at least one injury) = 1 – P (no injuries in 50 jumps) = 1 = (49/50) 50 = 0.636. Your friend is not correct. 2.133 Define the events: G : student guesses C : student is correct ) | ( C G P = ) 2 (. 25 . ) 8 (. 1 ) 8 (. 1 ) ( ) | ( ) ( ) | ( ) ( ) | ( + = + G P G C P G P G C P G P G C P = 0.9412. 3.38 Note that Y ~ Binom( n = 4, p = 1/3) a. p ( y ) = ( 29 ( 29 y y y - 4 3 2 3 1 4 , y = 0, 1, 2, 3, 4. b. P ( Y ≥ 3) = p (3) + p (4) = 8/81 + 1/81 = 9/81 = 1/9. c. E ( Y ) = 4(1/3) = 4/3. d. V ( Y ) = 4(1/3)(2/3) = 8/9 4.25 12 / 31 125 . 125 . ) ( 4 2 2 2 0 = + = dy y ydy Y E , . 6 / 47 125 . 125 . ) ( 4 2 3 2 0 2 2 = + = dy y dy y Y E So, V ( Y ) = 47/6 – (31/12) 2 = 1.16.

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4.91 Let Y = water demand in the early afternoon. Then, Y ~ Exp(β = 100) a. P ( Y > 200) = 2 100 / 200 100 1 - - = e dy e y = .1353. b. We require the 99 th percentile of the distribution of Y : P ( Y > 99 . φ ) = 100
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