A Z score is simply your score on a particular measure (i.e. height, weight, income, IQ) minus the average score on that measure (in a particular population), divided by the standard deviation in that population:

Z score = (score – average score)/standard deviation

So the average Z score by definition is 0.

On the IQ scale, where the average IQ (among white Americans or among all Americans) is often defined as 100 and the standard deviation is often defined as 15, someone with an IQ of 70 has a Z score of -2, someone with an IQ of 85 has a Z score of -1, someone with am IQ of 100 has a Z score of 0, someone with an IQ of 115 has a Z score of 1, someone with an IQ of 138 has a Z score of 2.5 etc.

Of course it depends who the reference group is.  If you’re comparing your IQ not to the general white population, but to white PhDs where the average IQ is about 123 and the standard deviation is about 13, then an IQ of 70 has a Z score of -4.07 and an IQ of 138 has a Z score of 1.15.

Why is this useful?

A good analogy is if you had 50 British pounds, 50 South African rands, and 50 Euros, and you wanted to know which of these three monies was worth most, you would convert them all to U.S. dollars, which serves as a kind of universal financial yardstick.

Analagously, Z scores are like a universal yard stick.  Let’s say a white man has a height of 200 cm, a weight of 200 lbs, and an IQ of 200.  He wants to know whether he should be become a basketball player where he needs height, a football player where he needs weight, or a scientist where he needs IQ.

If he knows that white men have a mean height of 179 cm (SD = 6.52 cm), a mean weight of 185 lbs (SD = 35 lbs) and a mean IQ of maybe 101 (SD = 15.5), then using the formula:

Z score = (score – average score)/standard deviation

he could calculate that relative to white men, he has a Z score of 3.2 when it comes to height, 0.42 when it comes to weight, and 6.4 when it comes to IQ.

All three Z scores are positive, so he is above the average for white men in all three traits, but he is most impressive in IQ and least impressive in weight, so he might decide to become a scientist and not a football player.  Further, if these three traits are normally distributed, he can apply this calculator.  By clicking the “one sided” option and then entering his Z scores, he would learn that for height (Z = 3.2), he’s taller than 99.9313% of white men, for weight (Z = 0.42), he’s heavier than 66.2757% of white men, and for IQ (Z = 6.4), he’s smarter than roughly 100% of white men.

Normalized Z scores

Z scores are most meaningful when traits are normally distributed.  When traits are not normally distributed they can often give silly results.  For example, Bill Gates net worth, at his peak, was \$100 billion dollars.  According to one recent study, the average net worth of a sample of Americans was \$145,837 with a standard deviation of \$447,814.

So Bill Gates financial Z score = (100,000000000 – 145,837)/447,814 = 223,306.67

Given that Z scores of 6 or higher are only supposed to occur one in a billion times, a Z score of 2,23306.67 is kind of meaningless.

A brilliant Promethean told me that when calculating the financial Z scores of billionaires, I should instead use normalized Z scores.  In other words, the financial Z score they would have if money were normally distributed.

So assuming there are 200 million American adults, the normal curve predicts the richest American adult (Bill Gates) should have a financial Z score of 5.73, so that would be his normalized Z score.

Z scores and correlations

If you know the correlation (r) of two variables, X and Y, then, assuming a bivariate normal distribution, you can estimate the average Z score on Y for people with a given Z score on X, by multiplying their Z score on X by r.

Conversely, you can estimate the average Z score on X for people with a given Z score on Y, by multiplying their Z score on Y by r.

So if the correlation between IQ and financial success is 0.4, people with a Z score of 2 on financial success should average a Z score of 2(0.4) = 0.8 on IQ, and people with a Z score of 2 on IQ, should also average a Z score of 0.8 on financial success.

The reason this happens is that a correlation can be thought of as a bunch of people plotted on a scatter plot, where each person’s IQ is the Y coordinate and each person’s income is the X coordinate (or vice versa as we see below): As you can see, in a scatter plot, there is a line of best fit,  which is just a line that comes closest to all the points on the scatter plot.  This line allows you to predict the average level of Y for a given level of X.

Now if both variables in the above scatter plot were expressed as Z scores, and a bivariate normal distribution is assumed, then the correlation between both variables can be thought of as the slope of the line of best fit.  Thus multiplying a given Z score on X by the correlation between X and Y allows you to deduce the average Z score on Y for said value of X (i.e. the Y coordinate for that value of X on the line of best fit).

Of course, traits like income are not normally distributed, particularly at the extremes, so trying to estimate the intelligence Z score of Bill Gates from his financial Z score of 223,306.67 by multiplying 223,306.67 by the IQ-income correlation of 0.4 would lead to an intelligence Z score of 89,322.67 and thus an IQ above one million!!!

Clearly that’s nonsense.  As a brilliant Promethean once told me, the correct procedure would be to use the normalized financial Z score of Bill Gates (5.73), and multiply that by the IQ-income correlation to get his expected IQ.

Of course, given that the IQ-income correlation is only 0.45, there would be an enormous amount of scatter around the line of best fit, so it would not be very accurate for a specific individual.

I once used a similar approach to estimate the IQ of Jesus (not from wealth, but from influence).  This was disputed by the highly respected commenter Misdreavous who wrote:

““Fame” (both academic and otherwise) is not normally distributed. Knowing that Jesus is the third most influential man alive (according to whom?) does not allow you to convert a rank order into a z-score. You would think this would be obvious to anyone who passed high school statistics.

However, if the approach was good enough for a brilliant Promethean, it is good enough for me.  Though I should note that the conventional method for normalizing non-Gaussian data is not to convert the data into normalized Z scores as I do, but instead to take the natural logarithm which I have not tried doing yet.