IQ Test Probability

Question

IQ Tests and scores are scaled to have a mean of 100 and a standard deviation of 15. They also are approximately normally distributed. A person scores a 125 on the IQ test. What is the percentage of people scoring higher than this person?

I'm having trouble calculating this answer. Please help. Thank you.

Answer 1

Given a normal distribution $f(x)$ with mean $100$ and standard deviation $15$, where $x$ is the IQ, the probability of a person having a higher IQ than $125$ is $$\int_{125}^\infty f(x)dx,$$

which is your answer. Can you fill in the details? (Note, however, that you won't get a closed form expression out of this, you'd have to use numerical methods.)

EDIT: In the comments above, you write "I am getting tripped up that $N$ (number of people) is not defined and I don't know what the maximum score of the test is?" When dealing with percentages, $N$ doesn't have to be defined; for a given $N$, $N$ times the probability you found is the expected number of people with an IQ above $125.$ Note that expected values do not have to be integers, even though it describes something that is discrete (number of people). With respect to the maximum score, it is infinity, as the range of the normal distribution is as such. This means that you, although extremely unlikely, have a non-zero chance of observing a person with an arbitrarily high IQ.

Answer 2

I'm pretty sure you are expected to use printed tables of the standard normal CDF to get a numerical solution to this problem. That involves transforming the IQ scale with $X \sim \mathsf{Norm}(\mu = 100, \sigma=15).$ To a standard normal scale with $Z \sim \mathsf{Norm}(0, 1).$ Essentially $Z$ is the number of standard deviations ($\sigma=15$) above or below the IQ mean ($\mu = 100$) that a particular IQ score falls (as in the Comment of JMoravitz).

Here is a sketch of the normal curve that describes IQ scores. The total area under the curve is $1 = 100\%.$ The probability you want is represented by area under the curve and to the right of the vertical red line. Roughly, it looks to be about 5%.

Here is how to begin a formal computation:

$$P(X > 125) = P\left(\frac{S - \mu}{\sigma} > \frac{125 - 100}{15} = 1.67\right) = P(Z > 1.67).$$

To use printed standard normal tables found in most elementary statistics texts, you will have to read the details because the formats of these tables vary. (Maybe you will have to find the area under the standard normal curve between 0 and 1.67 from that tables, and then subtract from 0.5.)

This problem can be solved directly (without standardization or rounding) using software. Using R statistical software, the computation is shown below.

 1 - pnorm(125, 100, 15)
 ## 0.04779035

You will likely not get exactly this result from printed normal tables, because it is necessary to round 1.6666... to 1.67 in order to use most tables.