document.write( "Question 961781: With a binomial distribution with n = 25 and p= 0.48, which is larger?\r
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document.write( "P(12 successes
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document.write( "P(9 successes
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document.write( "P(20 successes
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document.write( "P( 10 sucesses \n" );
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Algebra.Com's Answer #587654 by Edwin McCravy(20055)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "The mean of a binomial distribution where there are: \r\n" ); document.write( "\r\n" ); document.write( "n trials with a probability p of ONE sucess in ONE trial,\r\n" ); document.write( "\r\n" ); document.write( "is given by m = np. \r\n" ); document.write( "\r\n" ); document.write( "The nearer a number of successes is to the mean of m = np, \r\n" ); document.write( "the more likely it is to occur.\r\n" ); document.write( "\r\n" ); document.write( "In this case np = (25)(0.48) = 12.\r\n" ); document.write( "\r\n" ); document.write( "Therefore the most likely number of successes is 12.\r\n" ); document.write( "The next most likely is 10, the next is 9, and the least likely is 20.\r\n" ); document.write( "\r\n" ); document.write( "P(20 sucesses) < P(9 sucesses) < P(10 successes) < P(12 sucesses) \r\n" ); document.write( "\r\n" ); document.write( "FYI, by actual calculation,\r\n" ); document.write( "\r\n" ); document.write( "P(12 successes) = 0.15811\r\n" ); document.write( "P(9 successes) = 0.07897\r\n" ); document.write( "P(20 successes) = 0.00085\r\n" ); document.write( "P( 10 sucesses) = 0.11664\r\n" ); document.write( "\r\n" ); document.write( "The correct answer is, of course, P(12 successes).\r\n" ); document.write( "\r\n" ); document.write( "Edwin\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |