document.write( "Question 1085659: The probability of tossing heads with a weighted coin is 0.4. If this unfair coin is tossed
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Algebra.Com's Answer #699762 by Edwin McCravy(20059)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "First we see whether the normal approximation is valid by \r\n" );
document.write( "calculating np and n(1-p) to see whether they are both at \r\n" );
document.write( "least 5. np = (50)(0.4) = 20 and n(1-p) = (50)(1-0.4) =\r\n" );
document.write( "(50)(0.6) = 30.  So the approximation is valid.\r\n" );
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document.write( "\"mu+=+n%2Ap+=+%2850%29%280.4%29=20\"\r\n" );
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document.write( "\"sigma+=+sqrt%28np%281-p%29%29=sqrt%28%2850%29%280.4%29%281-0.4%29%29=3.464101615\"\r\n" );
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document.write( "The word \"inclusive\" follows the words \"between 15 and \r\n" );
document.write( "25 times\".  Therefore, we get the area under the normal curve \r\n" );
document.write( "from 0.5 less than the smaller value 15, that is, from x=14.5 \r\n" );
document.write( "to 0.5 greater than the greater value 25, that is, to x=25.5.\r\n" );
document.write( "We can do this either with technology or with a normal table.\r\n" );
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document.write( "With a TI-84\r\n" );
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document.write( "normalcdf(14.5,25.5,50*0.4,√(50*.4*0.6))\r\n" );
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document.write( "We get 0.8876487995\r\n" );
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document.write( "[Personally, I doubt that is any more accurate than 0.8 or 0.9]\r\n" );
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document.write( "Edwin
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