document.write( "Question 884672: The heights of 3000 women at a particular college are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches.
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\n" ); document.write( "How many women have a height between 62.5 and 67.5 inches?______ women \n" ); document.write( "
Algebra.Com's Answer #534602 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi
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document.write( "*Note: \"z+=+blue%28x+-+mu%29%2Fblue%28sigma%29\"
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document.write( "For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
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document.write( "Area under the standard normal curve to the left of the particular z is P(z)
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document.write( "Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50%  to the right
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document.write( "one  standard deviation from the mean accounts for about 68% of the set 
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document.write( "two standard deviations from the mean account for about 95%
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document.write( "and three standard deviations from the mean account for about 99.7%.\r
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document.write( "mean of 65 inches and a standard deviation of 2.5 inches
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document.write( "P(62.5 > x < 67.5) = P(-1 > z < 1) = ~.68 (Empirical Rule Above)
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document.write( "\"3000%2A.68\" = 2040 women have a height between 62.5 and 67.5 inches \n" );
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