The probability of getting someone with an IQ score of at least 133 is 0.0139.
Given that,
Membership in Mensa requires an IQ score above 131.5.
Nine candidates take IQ tests, and their summary results indicate that their mean IQ score is 133.
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
We have to determine,
If one person is randomly selected from the general population, find the probability of getting someone with an IQ score of at least 133.
According to the question,
Let X be the random variable that represents the scores of a population.
The probability of getting someone with an IQ score of at least 133 is,
The normal standard distribution and the z score are given by:
[tex]\rm z = \dfrac{x-\mu}{\sigma}\\\\[/tex]
Substitute all the values in the formula,
[tex]\rm P(X\geq 133)= P(\dfrac{x-\mu}{\sigma}) = P(Z\geq \dfrac{133-100}{15}) = P(Z\geq \dfrac{33}{15}) = P(Z\geq 2.2)[/tex]
The probability using the complement rule is,
[tex]\rm P(X\geq 133) = 1-P(Z<2.2) = 1-.986= 0.0139[/tex]
Hence, The probability of getting someone with an IQ score of at least 133 is 0.0139.
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https://brainly.com/question/13846127
