Question

0. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Mensa is an international society that has one and only one qualification for membership: a score in the top 2% on an IQ test. a. What IQ score should one have in order to be eligible for Mensa?________ b. In a typical region of 115,000 people, how many are eligible for Mensa?

Answer 1

Answer:

a) $a=100 +2.05*15=130.75$

So the value for the scores that separates the bottom 98% of data from the top 2% is 130.75.  

b) For this case since just 2% of the people satisfy the condition then the number of expected people satisfying the requirement are:

$N = 115000 *0.02 =2300$

Step-by-step explanation:

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".  

Part a

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:

$X \sim N(100,15)$  

Where $\mu=100$ and $\sigma=15$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.02$   (a)

$P(X<a)=0.98$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.98 of the area on the left and 0.02 of the area on the right it's z=2.05. On this case P(Z<2.05)=0.98 and P(z>2.05)=0.02

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.98$  

$P(z<\frac{a-\mu}{\sigma})=0.98$

But we know which value of z satisfy the previous equation so then we can do this:

$z=2.05<\frac{a-100}{15}$

And if we solve for a we got

$a=100 +2.05*15=130.75$

So the value for the scores that separates the bottom 98% of data from the top 2% is 130.75.  

Part b

For this case since just 2% of the people satisfy the condition then the number of expected people satisfying the requirement are:

$N = 115000 *0.02 =2300$