Question 1161585: The average IQ is 100. The distribution is normal with a standard deviation of 15.
A. What is the probability that a randomly selected person has an IQ greater than 117?
B. What percentage of the population has an IQ between 88 and 108
C. What IQ represents the 98th percentile? (That’s the score that 98% of the population would fall below.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd = (117-100)/15 =17/15 or 1.133
probability z>1.133 is 0.1285
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%.
for the 98th percentile, which has a z-value of >2, the IQ should be greater than 130
use invnorm (2 VARS 3) (.98, 100, 15) for 130.81
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