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2 answers:
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Answer:
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
And the best way to solve this problem is using the normal standard distribution and the z score given by:
And we can find the following probablity:
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