IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. If a certain statistician has an IQ of 112, what percent of the population has an IQ less than she does?
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Doing a z-test, therefore z<((112-100)/15) This gives you your z-value, and so look in the normal tables for Z<0.133333, and you will find that it is P=0.55303 or 55% which ever you prefer, which is the percentage of people who have an IQ less than 112, if it was greater then it would be P=1-0.55303. I hope this helps.
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