# Statss - won by the manximum number of customers which is...

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Julio C Lopez MAST 6201 3/28/2016 Homework Session #1 p = .05 Falls = 0 = 1-p Var = .25 StDev= .5 SQRT(p(1-p)) Weak Team SQRT(p(1-p)) var = .1875 Wins= .75 = 1 Wins=.25=1 Loses= .25 =0 Loses=.75=0 StDev = .433 StDev = .433 3.) 5.1 Distribution A a. ExpVal = 1 (0*.5)+(1*.02)+(2*.15)+(3*.1)+(4*.05) b. StDev = 1.22 Distribution B a. ExpVal = 3 (0*.5)+(1*.01)+(2*.15)+(3*.02)+(4*.05) b. StDev = 1.22 4.) 5.3 b. ExpVal = \$5.49 15000*(1/31478)+ 500*(1/31478)+ 5*(31476/31478) c. StDev = \$84.56 sqrt ((5-5.49)^2*(31476/31478)+(500-5.49)^2*(1/31478)+(15000-5.49)^2*(1/31478)) 1.) Rises =1 = p 2.) Top Team SQRT((0-1)^2*0.5+(1-1)^2*0.2+(2-1)^2*0.15+(3- 1)^2*0.1+(4-1)^2*0.05) SQRT((0-3)^2*0.05+(1-3)^2*0.1+(2-3)^2*0.15+(3- 3)^2*0.2+(4-3)^2*0.5) c. The expected value for Distribution B is higher than that of Distribution A ( 3 > 1) but the StDev are exactly the same. a. The probability for the first 2 are 1/31,478 and the 3rd is 31476/31478, the number of fliers sent out by the autombile dealer = 31,478 d. Yes, I think the promotion was seems very effective because the expected value of the prize won is relatively close to the \$5 of the actual prize

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Unformatted text preview: won by the manximum number of customers, which is 31,476. So the pay out is so small that with selling a few more cars than usual will more than o³set the cost. The standard devia±on is coming out to be large because the varia±on of the amounts of the other 2 prizes skews it. Julio C Lopez MAST 6201 3/28/2016 5.) 5.5 a. ExpVal = 2.9 1*(31/200)+2*47/200)+3*(41/200)*(29/200)+5*(21/200)+6*(10/200)+7*(5/200)+8*(2/200) b. StDev= 1.772 6.) 5.21 p = .5 0.03125 0.5^5 x = 5 n = 5 7.) Skip 8.) 5.23 a. 0.000976563 BINOM.DIST(5,5,0.25,FALSE) b. 0.015625 BINOM.DIST.RANGE(5,0.25,4,5) c. 0.23730469 BINOM.DIST(5,5,0.75,FALSE) d. 0.89648438 BINOM.DIST.RANGE(5,0.25,0,2) 9.) 5.25 a. ExpVal = 1 20*.05 StDev = .974679434 SQRT(20*0.05*0.95) b. 0.358485922 BINOM.DIST(20,20,0.95,FALSE) c. 0.377353603 BINOM.DIST(1,20,0.05,FALSE) d. 0.2641605 BINOM.DIST.RANGE(20,0.05,2,20) SQRT((0-2.9)^2*(14/200)+(1-2.9)^2*(31/200)+(2-2.9)^2*(47/200)+(3-2.9)^2*(41/200)+(4-2.9)^2*(29/200)+(5-2.9)^2*(21/200)+(6-2.9)^2*(10/200)+(7-2.9)^2*(5/200)+(8-2.9)^2*(2/200))...
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