Answer:
![P(Z>3) = 1-P(Z](https://tex.z-dn.net/?f=%20P%28Z%3E3%29%20%3D%201-P%28Z%3C3%29%3D%201-0.99865%3D0.00135)
So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135
And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025
And we can say that the answer is a.2
Step-by-step explanation:
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Solution to the problem
Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:
We are interested on this probability
![P(X>X+3\mu)](https://tex.z-dn.net/?f=P%28X%3EX%2B3%5Cmu%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
And we can find the following probablity:
![P(Z>3) = 1-P(Z](https://tex.z-dn.net/?f=%20P%28Z%3E3%29%20%3D%201-P%28Z%3C3%29%3D%201-0.99865%3D0.00135)
So then the probability that an individual present and IQ higher than 3 deviation from the mean is 0.00135
And if we find the number of individuals that can be considered as genius we got: 0.00135*1500=2.025
And we can say that the answer is a/2.0