document.write( "Question 1161585: The average IQ is 100. The distribution is normal with a standard deviation of 15.\r
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document.write( " A. What is the probability that a randomly selected person has an IQ greater than 117?\r
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document.write( " B. What percentage of the population has an IQ between 88 and 108\r
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document.write( " C. What IQ represents the 98th percentile? (That’s the score that 98% of the population would fall below.)
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Algebra.Com's Answer #785202 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(x-mean)/sd = (117-100)/15 =17/15 or 1.133 \n" ); document.write( "probability z>1.133 is 0.1285\r \n" ); document.write( "\n" ); document.write( "between 88 and 108 one can use the calculator and do 2nd VARS 2 Normcdf (88,108,100,15) to get a probability that is 0.4912 or 49%.\r \n" ); document.write( "\n" ); document.write( "for the 98th percentile, which has a z-value of >2, the IQ should be greater than 130 \n" ); document.write( "use invnorm (2 VARS 3) (.98, 100, 15) for 130.81 \n" ); document.write( " |