HW6_2019.pdf - ENGG2470A Probability and Statistics for Engineers Spring 2019 Solution for Assignment 6 Date Apr 17th 2019 Prob 13 Solution(20pts Let Xi

HW6_2019.pdf - ENGG2470A Probability and Statistics for...

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ENGG2470A: Probability and Statistics for Engineers Spring 2019 Solution for Assignment 6 Date: Apr 17th, 2019 Prob. 13. Solution: (20pts) Let X i be the number of pennies added on day i. we have EX i = 3 . 5 and V ar ( X ) = 35 12 . Let Y n = n i =1 X i , which is the total number of pennies addd on/before day n. We have Y n = 3 . 5 n and V ar ( Y n ) = 35 12 × n . Then event that it takse at least 80 days to collect 3 dollars is equivalent to { Y 79 < 300 } Pr ( Y 79 < 300) = Pr ( Y 79 - 3 . 5 × 79 q 35 12 × 79 < 300 - 3 . 5 × 79 q 35 12 × 79 ) (1) Φ( 300 - 3 . 5 × 79 q 35 12 × 79 ) (2) = Φ(1 . 548) (3) = 0 . 9392 (4) Some of you calculate Pr { Y 79 299 } . It’s totally fine. Pr { Y 80 ( < )300 } is not so rigorous, but will be regarded as right as well. Prob. 14. Solution: (20pts) We have EX i = 1 2 , V ar ( X i ) = 1 12 , i = 1 , 2 ... 10 Since X i are independent,let X = 10 i =1 X i , then EX = 10 EX i = 5, V ar ( X ) = 10 V ar ( X i ) = 5 6 (a) P ( X 7) EX 7 = 5 7 (b) P ( X 7) = P ( X - 5 7 - 5) = 1 2 P ( | X - 5 | ≥ 2) 1 2 V ar ( X ) 2 2 = 5 6 × 8 = 0 . 104 (c) P ( X 7) = P ( X - 5 V ar ( X ) 7 - 5 V ar ( x ) ) 1 - Φ( 7 - 5 V ar ( x ) ) = 1 - Φ(2 . 19) = 0 . 0143 Prob. 15. Solution: (20pts) S n has mean n 2 ,variance 1 4 n According to CLT, when n is large, S n - n 2 1 2 n follows N (0 , 1).
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(a) as n → ∞ P ( n 2 - 10 S n n 2 + 10) =
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Unformatted text preview: P (-10 1 2 √ n ≤ S n-n 2 1 2 √ n ≤ 10 1 2 √ n ) = 2Φ( 10 1 2 √ n )-1 → (b) as n → ∞ P ( n 2-n 10 ≤ S n ≤ n 2 + n 10 ) = P (-1 10 ≤ S n n-1 2 ≤ 1 10 ) → 1 by weak law of large number. (c) as n → ∞ P ( n 2-√ n 2 ≤ S n ≤ n 2 + √ n 2 ) = P (-1 ≤ S n-n 2 1 2 √ n ≤ 1) → 2Φ(1)-1 = 0 . 6827 Prob. 4. Solution: (20pts) (a) X i are independent (b) X i are iid with mean 0 and variance 1 12 , let Y n = ∑ n i =1 X i , then EY n = 0 , V ar ( Y n ) = n 12 p = P ( | Y n | ≥ c ) = P ( | Y n √ n 12 | ≥ c √ n 12 ) ≈ 2(1-Φ( c √ n 12 )) Here c = 10 , n = 100,then p = 2(1-Φ( 10 √ 100 12 )) = 0 . 054% (c) No, according to CLT. Prob. 5. Solution. (20pts) var ( X ) = var ( ˆ X + Δ) = var ( ˆ X ) + var (Δ) + 2 cov ( ˆ X, Δ) = var ( ˆ X ) + var (Δ) 2...
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