document.write( "Question 659927: I need HELP: Given a binomial distribution with n=20 and p=0.76, would the normal distribution provide a reasonable approximation? Why or Why Not?
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\n" ); document.write( "Now is this the type of problem where the n represents the number of tries?
\n" ); document.write( "like this
\n" ); document.write( "n=20 tries
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\n" ); document.write( "or something like that? Please help me....... \n" ); document.write( "
Algebra.Com's Answer #411003 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Some books say:\r\n" );
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document.write( "Calculate np and n(1 - p) and if both are\r\n" );
document.write( "greater than 5, the approximation is good.\r\n" );
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document.write( "np > 5 and n(1 - p) > 5\r\n" );
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document.write( "np = 20(0.76) = 15.2\r\n" );
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document.write( "n(1 =- p) = 20(1 - 0.76) = 4.8\r\n" );
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document.write( "The second one is not > 5 so it would not be a good approximation.\r\n" );
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document.write( "Other books say:\r\n" );
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document.write( "Calculate np(1 - p) and if this is >= 10 then the approximation is good.\r\n" );
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document.write( "np(1 - p) = (20)(0.76)(1 - 0.76) = 3.648\r\n" );
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document.write( "That is not greater than or equal to 10\r\n" );
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document.write( "So the normal approximation would not be good in either case.\r\n" );
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document.write( "Edwin
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