Question

Question #fb52f

Answer 1

A) `~~75`
B) `~~25.3%`
C) `1.33` (two decimals)

Explanation:

To answer these questions, we would have to know if the standard IQ test results fall into a normal distribution. There is some debate online about that (many say it's meant to be a normal distribution, while the actual results are more likely to diverge from a normal distribution in the "tails"), but for this problem I believe it is intended to be treated that way.

Thus, this problem describes a normal distribution with a mean `mu` of 100 and a standard deviation `sigma` of 15. In notation terms, we say this is a `N(100,15^2)` distribution. To answer the questions it is helpful to convert this problem into the equivalent standard normal distribution `N_s(0,1^2)` (mean of 0, std. deviation of 1) and then we will be able to reference standard normal distribution area/probability tables.

Question A

Asking what IQ score has 95% of people above it is precisely the same as asking what IQ score has 5% of people below it, since we know that all IQ scores combined would represent 100% of all scores. For the standard normal distribution, we need to know which z-score has a left-tail area that is 0.05; this would be the point where 5% (ie 0.05) of the area under the standard normal distribution lies to the left of that z-score.

To get this, we can either consult a z-score table or use a z-score calculator. I will opt to use a z-score calculator, which finds that a z-score of -1.64 (to two decimals) is our target.

Now, we use the z-score formula to determine the correlated IQ score `x` according to our `N(100,15^2)` distribution this z-score matches:

`z = (x-mu)/sigma`

`-1.64 = (x-100)/15`

`-24.6 = x-100`

`x = 75.4~~75`

Question B

To find the percentage of people who would score less than 90, we work in an opposite direction than Question A. In this case, we will convert the IQ score of `x = 90` into a standard normal distribution z-score, and then consult a z-score calculator or table to determine the proportion of all z-scores that are lower than that value.

`z = (x-mu)/sigma`

`z = (90 - 100)/15`

`z = -10/15 = -0.67` (two decimals)

A z-score calculator tells us that for this z-score, approximately 25.3% of all z-scores lie to the left of this z-score. Another way of looking at this is saying that 25.3% of people would be expected to have a lower IQ score than 90.

Question C

This question is more straightforward. We can use the z-score formula to convert an IQ score of 120 from a distribution of `N(100,15^2)`:

`z = (x - mu)/sigma`

`z = (120 - 100)/15`

`z = 20/15 = 1.33` (two decimals)