What is the proportion of the population listed below is highly advanced (greater than or equal to 145?)

Question

Intelligence quotients (IQs) measured on the Standford Revision of the Binet-Simon Intelligence Scale are normally distributed with a mean of 100 and a standard deviation of 16.

IQ Range    IQ Classification
145-160     Highly Advanced
130-144     Gifted
120-129     Superior
110-119     Elevated
90-109      Average
80-89       Low average
70-79       Borderline impaired
55-69       Mildly impaired
40-54       Moderately impaired

Please help me with this, I am really confused.

Answer 1

Maybe this will put you on the right path;

Your threshold (145) minus the mean of the normal distribution (100) will give you 45. 45 divided by your standard deviation (16) gives you 2.8125 standard deviations.

There is some percentage of the population that falls outside of 2.8125 std devs, and since this is a normal distribution, half of those are in the range you're looking for, and half are on the other end of the distribution.

Answer 2

First, it is important to note that the proportion of the population with an IQ above or equal to 145, will be equivalent to the whole population (100%) minus those who have IQs of less than 145.

This fact is useful, because the cumulative distribution function for a given distribution gives you the proportion of values falling below a threshold, not above.

Therefore, our first step is to convert our distribution of IQs to the standard normal distribution to make it easier to work with. The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is useful to put our IQ threshold in terms of the standard normal distribution as its function is easy to work with.

The way we can convert our current threshold, 144, into an equivalent threshold in the standard normal distribution is by turning it into a z-score. The z-score is given by our threshold minus the distribution's mean over the distribution's standard deviation: (144-100)/16=2.75.

Now that we our threshold in terms of the standard normal distribution, we can plug it into our cumulative distribution function (CDF), which will tell us what proportion of the population have an IQ of 144 or less. The CDF is given by:

In this notation, our z-score of 2.75 will take the place of x in the integral. Essentially, we'll be calculating the area under the normal distribution curve from negative infinity to our threshold of 2.75 since we know that the total area from negative infinity to infinity is 1 (100% of the population).

The result of plugging in our z-score is 0.99702, and 1-0.99702=0.00298, which is 0.298%.

One can avoid having to use the actual CDF by making use of the statistical functions in excel. Using the =norm.dist(threshold, mean, standard deviation, cumulative=1) function, excel will do the conversion into the standard normal distribution and give you the same result.