Assignment 4 complete

# Assignment 4 complete - Mir Jamiur Rahman Student number...

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Mir Jamiur Rahman Student number: 212115341 MGMT 1050 Section D 1. Variance= 135.5-(9.65)² = 42.3775 Standard Deviation = √(42.3775)= 6.5098 Ans. Mean= 9.65 Standard Deviation= 6.5098 2. Variance= 1.15- (1)² = 0.15 Ans. Mean= 1 Variance= 0.15 3. E(X)= (0x0.1+1x0.25+2x0.4+3x0.2+4x0.05) = 1.85 Profit= 20% of \$20 = 20x(0.20) = 4 4x1.85= 7.4 Ans. Expected value of profit contribution is \$7.4 4. n= 25 p= 10%

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Mir Jamiur Rahman Student number: 212115341 MGMT 1050 Section D a) P(None of the people are independent)= 0.0718 (Using tables) b) P(Fewer than 5 people are independent)= 0.9020 (Using tables) c) P(More than 2 are independent)= 1-0.5371= 0.4629 (Using tables) 5. N= 10 P= 244/495 a) P(less than 4 in 10 games): =BINOM.DIST(4,10,244/495,1) = 0.39447 P(Winning 5 or more times in 10 games)= 1- 0.39447= 0.60553 b) Expected wins in 100 games= np= 100x(244/495)= 49.2929 6. n= 25 p(Red)= 18/38 p(Black)= 18/38 p(Green)= 2/38 a) P(Ball falls in green slots 1 or less times): =BINOM.DIST(1,25,2/38,1) = 0.618256 P(Ball falls in green slots 2 or more times)= 1-0.618256= 0.381744 b) P(Ball does not fall into any green slots): =BINOM.DIST(0,25,2/38,1) = 0.258805 c) P(Ball falls into black slots 14 or less times): =BINOM.DIST(14,25,18/38,1) = 0.856449 P(Ball falls into black slots 15 or more times)= 1-0.856449= 0.143551 d) P(Ball falls into red slots 10 or fewer times): =BINOM.DIST(10,25,18/38,1) = 0.296796
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• Fall '12
• HilaCohen;OlgaKaminer;AlanMarshall;AlexShoumarov,DoritNevo
• Standard Deviation, 10%, \$20, Cauchy distribution, \$7.4 4

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