Question

For a given IQ test, an individual is considered a genius if their score falls more than three standard deviations from the mean. Assuming the population of 1500 individuals fits a normal deviation, how many individuals can you expect to be considered a genius?

a. 2
b. 5
c. 3
d. 8
e. 10
f. 1

Answer 1

Answer:

$P(Z>3) = 1-P(Z<3)= 1-0.99865=0.00135$

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:

$X \sim N(\mu=?,\sigma=?)$  

We are interested on this probability

$P(X>X+3\mu)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

And we can find the following probablity:

$P(Z>3) = 1-P(Z<3)= 1-0.99865=0.00135$

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