document.write( "Question 909190: 7% of the customers did not receive delivery \r
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document.write( "If a sample of 50 customers is selected on a given day, assuming that the trials are independent of each other with one outcome, what is the probability that. \r
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document.write( "a. fewer than 3 customers would receive a free newspaper?\r
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document.write( "b. 2, 3, or 4 customers would receive a free newspaper? \r
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document.write( "c. more than 5 customers would receive a free newspaper? \r
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document.write( "I believe n=50 P=.07 and X=3 for part A\r
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document.write( "I have tried this problem multiple times using the Binomial probability formula, but all my answers seem incorrect. \n" );
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Algebra.Com's Answer #551663 by swincher4391(1107)![]() ![]() You can put this solution on YOUR website! X~binomial(n=50, p=.07)\r \n" ); document.write( "\n" ); document.write( "A) P[X<3] = P[X<=2] = P[X=0] + P[X=1] + P[X=2] = (50 choose 0)(.07)^0(.93)^50 + (50 choose 1)(.07)^1(.93)^49 + (50 choose 2)(.07)^2(.93)^48 = .3108\r \n" ); document.write( "\n" ); document.write( "B) P[X=2] + P[X=3] + P[X=4] = (50 choose 2)(.07)^2(.93)^48 + (50 choose 3)(.07)^3(.93)^47 + (50 choose 4)(.07)^4(.93)^46 = .6025\r \n" ); document.write( "\n" ); document.write( "C) P[X>5] = 1-P[X<=5] (using the complement). \r \n" ); document.write( "\n" ); document.write( "1-(P[X=0] + P[X=1] + (P[X=2] + P[X=3] + P[X=4])+ P[X=5])\r \n" ); document.write( "\n" ); document.write( "1-(P[X=0] + P[X=1] + B + P[X=5]) where B is the answer from part B.\r \n" ); document.write( "\n" ); document.write( "1-((50 choose 0)(.07)^0(.93)^50 + (50 choose 1)(.07)^1(.93)^49 + .6025 + (50 choose 5)(.07)^5(.93)^45) = .1351\r \n" ); document.write( "\n" ); document.write( "Hope this helps!\r \n" ); document.write( "\n" ); document.write( "Devin [swincher4391@yahoo.com]\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |