Question

IQ scores (as measured by the Stanford-Binet intelligence test) in a certain country are normally distributed with a mean of 100 and a standard deviation of 18. Find the approximate number of people in the country (assuming a total population of 323,000,000) with an IQ higher than 126. (Round your answer to the nearest hundred thousand.)

Answer 1

This is a normal distribution question with

P(x > 126.0)=?
The z-score at x = 126.0 is,

z = 1.4444
This implies that
P(x > 126.0) = P(z > 1.4444) = 1 - 0.9257

approximate number of people in the country with IQ higher than 126 = total population * P(x > 126)

approximate number of people in the country with IQ higher than 126 = 323000000 * 0.0743

approximate number of people in the country with IQ higher than 126 = 23998900

PS: you have to refer z score table to find the final probabilities.
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