Assignment 6 solution

Assignment 6 solution - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT0301 Elementary Statistical Methods Semester 1 2009/2010 ASSIGNMENT 6 SOLUTIONS Q1 . The profits table: Demand\stock 250 350 450 prob 250 ($20less$12)x250= $2000 ($8x250)less($12x100)= $800 ($8x250)less($12x200)= -$400 0.2 350 $8x250= $2000 $8x350= $2800 ($8x350)less($12x100)= $1600 0.4 450 $8x250= $2000 $8x350= $2800 $8x450= $3600 0.4 Expected profit $2000x(0.2+0.4+0.4)= $2000 $800x0.2+$2800x(0.4+0.4)= $2400 (minus$400x0.2)+($1600x0.4)+($3 600x0.4)= $2000 tot= 1 The merchant should stock 350 items as to maximize his expected daily profit. The maximum expected profit is $2400.
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Q2. Group Male(M) Female(F) Drinker(D) 30 20 Smoker(S) 19 5 Both(B) 8 0 Otherwise(O) 49 35 total F=60 D S O= 35 total M=90 D S O= 49
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(a) Because 150 24 150 5 19 ) ( and 150 50 150 20 30 P(D) , 150 8 ) ( = + = = + = = S P S D P and ) ( P(D) ) ( S P S D P = The events of this member being a drinker and being a smoker are independent. (b)
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This note was uploaded on 04/14/2010 for the course STAT 0301 taught by Professor Gabriell during the Spring '09 term at HKU.

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Assignment 6 solution - THE UNIVERSITY OF HONG KONG...

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