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The IQ of adults is normally distributed with a mean of 100 and a standard deviation of 15. To determine the probability that that an adult has an IQ greater than 131.5, find the z-score of 131.5
Here `mu = 100` and `sigma = 15` .
The z-score is `(131.5 - 100)/15 = 2.1`
Use a normal distribution table to find the corresponding area for this z-score. The value is 0.9821. 98.21% of values lie below 131.5. The probability that an adult has an IQ greater than 131.5 is 1 - 0.9821 = 0.0179
There is 1.79% probability that the IQ of a randomly chosen adult is greater than 131.5
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