PS4s - X + σ X ) = P (2 . 2111 < X < 5 . 7889) = P (...

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SOLUTIONS TO ISMT111 PROBLEM SET #4 (1) Answer: 0.002. Binomial n = 25, p = 0 . 7 for nondefective. P ( X 10) = 0 . 002 from Table. (2) A Binomial Problem (a) The expected number of applicants who will pass the test and the standard deviation are, respectively,: Answer: 6 and 2.049. Binomial np = (20)( . 3) = 6 and npq = p (20)( . 3)( . 7) = 2 . 049 . (b) The probability that between 5 and 10 applicants inclusively will pass the test is: Answer: 0.745. (= 0 . 983 - 0 . 238) from table. (3) Answer: 0.7582 A hypergeometric problem: ( 10 2 )( 5 1 ) ( 15 3 ) + ( 10 3 )( 5 0 ) ( 15 3 ) = 345 455 (4) Answer: 0.5980 X is binomial with n = 20 and p = 0 . 20. μ X = np = 4 , σ X = 3 . 2 = 1 . 7889. P ( - σ X < X - μ X < σ X ) = P ( μ X - σ X < X < μ
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Unformatted text preview: X + σ X ) = P (2 . 2111 < X < 5 . 7889) = P ( 3 ≤ X ≤ 5) because X is an integer , = 0 . 804-. 206 = 0 . 598 (5) Answer: 4. A binomial problem with p = 0 . 9 and a sample size n =? such that 1-± n ² (0 . 9) (0 . 1) n-± n 1 ² (0 . 9) 1 (0 . 1) n-1 ≥ . 99 Check that if n = 3 , LHS = 0 . 9720 , and that if n = 4 , LHS = 0 . 9963 (6) Both methods should produce identical answers: Average Portfolio Return = 0.03709000; Standard Deviation of Portfolio Return = 0.070888009. * * * 1 ATChan 2008...
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This note was uploaded on 12/22/2009 for the course ISOM ISOM111 taught by Professor Anthonychan during the Fall '09 term at HKUST.

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