03exS_1-3

03exS_1-3 - N = 10 + 90 = 100 , D = 10 , n = 10 (a) Let the...

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ISOM111 Business Statistics (Fall 2010) Suggested Solutions for Tutorial Exercise 3 Please note that the tutorial exercise solutions may not be complete answers. Answers without showing the work in exam may earn you only partial credit. 1. (a) Mean of Stock X =$72 , 000 S.D. of Stock X =$80 , 099 . 9 Mean of Stock Y =$66 , 000 S.D. of Stock Y =$24 , 576 . 4 (b) The covariance between Stock X and Stock Y is 1752 , 000 , 000 The returns of Stock X and Stock Y are negatively correlated. (c) Let the return of this investment portfolio be P 1 μ P 1 = w x μ x + w y μ y =0 . 6 $72 , 000 + 0 . 4 $66 , 000 = $69 , 600 σ P 1 = p w 2 x σ 2 x + w 2 y σ 2 y +2 w x w y σ xy =$39 , 565 . 5 2. (a) Let the total cost of checking the f rst possibility f rst be X I X I 100 + 1000 100 + 500 + 2000 P ( X I ) 0 . 2 0 . 8 E [ X I ]=0 . 2($1100) + 0 . 8($2600) = $2300 (b) Expected total cost of checking the second possibility ( X II ) f rst: E [ X II ]=0 . 2($500 + $100 + $1000) + 0 . 8($500 + $2000) = $2320 E [ X I ] = $2300 <E [ X II ] = $2320 As the expected total cost of checking the f rst possibility f rst is lower, we should check the f rst possibility f rst. 1
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Unformatted text preview: N = 10 + 90 = 100 , D = 10 , n = 10 (a) Let the number of defective items in the sample be X and X ∼ Hyper (10 , 10 , 100) It is a hypergeometric random variable with n = 10 , S = 10 , N = 100 i. 2 possible outcomes: defective items or non-defective items ii. f nite population without replacement, i.e. dependent trials, p is not a constant (b) The expected number of defective items in the sample is E ( X ) = n S N = 10 ∗ 100 1000 = 1 Its standard deviation is σ : r n S ( N − S ) N 2 N − n N − 1 = r 10 ∗ 10(100 − 10) 100 2 ∗ 100 − 10 100 − 1 = √ . 818182 = 0 . 904534 (c) The sample is considered to be unacceptable if it contains more than two defective items P ( x > 2) = 1 − P ( x 6 2) P ( x 6 2) = P ( x = 0) + P ( x = 1) + P ( x = 2) = 2 X x =0 C 10 x C 90 10 − x C 100 10 = 0 . 93998 P ( x > 2) = 1 − . 93998 = 0 . 06002 2...
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This note was uploaded on 01/06/2011 for the course ISOM 111 taught by Professor Hu,inchi during the Fall '10 term at HKUST.

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03exS_1-3 - N = 10 + 90 = 100 , D = 10 , n = 10 (a) Let the...

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