document.write( "Question 926296: Intelligence quotients on the Stanford binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. Use the 68-95-99.7 rule to find the percentage of people with IQs
\n" ); document.write( "a)between 68 and 132
\n" ); document.write( "b)between 68 and 100
\n" ); document.write( "c)above 116
\n" ); document.write( "d)below 68
\n" ); document.write( "e)above 148
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Algebra.Com's Answer #562145 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.
\n" ); document.write( "Area under the standard normal curve to the left of the particular z is P(z)
\n" ); 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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\n" ); document.write( "......
\n" ); document.write( "one standard deviation from the mean accounts for about 68% of the set
\n" ); document.write( "two standard deviations from the mean account for about 95%
\n" ); document.write( "and three standard deviations from the mean account for about 99.7%.
\n" ); document.write( ".......
\n" ); document.write( "mean of 100 and a standard deviation of 16\r
\n" ); document.write( "\n" ); document.write( "a)between 68 and 132: (2SDs on either side of mean) 95%
\n" ); document.write( " b)between 68(-2SDs) and 100: 50% - 95%/2
\n" ); document.write( " c)above 116 (1SD): 50% + 68%/2
\n" ); document.write( " d)below 68(-2SDs): (100%-95%)/2
\n" ); document.write( " e)above 148 (3Sds): (100%-99.7%)/2
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