document.write( "Question 293233: Probability of getting at least 65 girls in 100 births \n" ); document.write( "
Algebra.Com's Answer #211738 by Edwin McCravy(20064)\"\" \"About 
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Probability of getting at least 65 girls in 100 births
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document.write( "This is a binomial probability.  We use the normal approximation\r\n" );
document.write( "of the binomial:\r\n" );
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document.write( "\"mu=np=100%2A0.5=50\"\r\n" );
document.write( "\"sigma=sqrt%28npq%29=sqrt%28100%2A0.5%2A0.5%29=sqrt%2825%29=5\"\r\n" );
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document.write( "Since the probability we want is P(x\"%22%22%3E=%22%22\"65), for\r\n" );
document.write( "proper estimation we use P(x\"%22%22%3E=%22%22\"64.5). \r\n" );
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document.write( "We find the z-score for x=64.5\r\n" );
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document.write( "\"z=%28x-mu%29%2Fsigma=%2864.5-50%29%2F5=2.90\"\r\n" );
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document.write( "We find 2.9 in the leftmost column for z and 0 across the top\r\n" );
document.write( "which means we want the entry just to the right of 2.9 which \r\n" );
document.write( "is .4981.  This is the area between z=0 and z=2.90, so to get\r\n" );
document.write( "the area right of that we must subtract 0.4981 from 0.5, and\r\n" );
document.write( "get .0019.\r\n" );
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document.write( "If we use a TI-83 or TI-84 we get \r\n" );
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document.write( "1-binomcdf(100,.5,64)=1-.9982411791=.0017588209\r\n" );
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document.write( "which rounds to .0018 rather than .0019 but we can expect such \r\n" );
document.write( "small differences between answers found using a graphing \r\n" );
document.write( "calculator and those found using a normal table.\r\n" );
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document.write( "Edwin
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