I have read several explanations of standard deviation and z-score, I know how to it calculate them but I am not sure what is differences betwen them.

Can someone explain it to me? When is suitable to use stdev and when z-score?

  • 3
    $\begingroup$ From here it follows that whenever $\mu$ and $\sigma$ are the mean and std. dev. of some distribution (hence, constants) then $z(x) = \frac{x-\mu}{\sigma}$ is the $z$-score of $x$ which depends on $x$ and defined using $\sigma$. In such case, $\sigma$ can be considered as a "unit" of measurements and $z(x)$ tells us on how many units is $x$ far from the mean $\mu$ $\endgroup$
    – Ilya
    Jan 11, 2013 at 14:04
  • $\begingroup$ @Ilya That's an answer, not a comment :) $\endgroup$
    – rschwieb
    Jan 11, 2013 at 14:05
  • $\begingroup$ @rschwieb: Thanks! I'm not a perfect statistician and only read a related wikipedia article now, so maybe someone more experienced is willing to leave a more comprehensive answer. If not, I'll maybe post my comment as an answer. $\endgroup$
    – Ilya
    Jan 11, 2013 at 14:07

1 Answer 1


Typically you have a collection of values of some variable. If the variable is human height, these values might be expressed in inches or centimetres; if it’s human weight, in pounds or kilograms. The standard deviation is a specific number of these units; roughly speaking, it’s the size of a typical deviation from the mean of these measurements. To compare two different collections of measurements, it’s generally very desirable to express them in units that make these typical deviations the same size. We might call such units standard units: standard units are units chosen so that the mean (average) of the measurements is $0$, and a typical deviation $-$ technically, the standard deviation $-$ has size $1$. The $z$-scores are just the original measurements expressed in these standard units instead of the original units of measurement.

The conversion is actually quite similar to the conversion between two more familiar units of measurement. If you measure a length in inches (or centimetres) and then convert that measurement to feet (or metres), you have to multiply by $\frac1{12}$ (or $\frac1{100}$), because each inch is $\frac1{12}$ of a foot (each centimetre is $\frac1{100}$ of a metre). Suppose that you’re measuring human heights, and you get a standard deviation of $3$ inches. Then for this sample $1$ standard unit is $3$ inches, and each inch is $1/3$ of a standard unit. Someone whose height is $4.5$ inches above the mean is $4.5\cdot\frac13=1.5$ standard units above the average. This deviation of $1.5$ standard units is the $z$-score corresponding to that height of $4.5$ inches above average.

You might say that the standard deviation is a yardstick, and a $z$-score is a measurement expressed in terms of that yardstick.

The situation is a little different from simple conversion between inches and feet, though. It’s more like conversion between Fahrenheit and Celsius temperatures: not only does the size of the unit change, but also the $0$ point. Just as $0^\circ$ C is $32^\circ$ F, not $0^\circ$ F, the $0$ point for $z$-scores is generally not the same as the $0$ point for the actual measurements. A sample of adult American males, for instance, might have an average height of $70$ inches, so that someone $4.5$ inches above the average would be $74.5$ inches tall. The $0$ point for $z$-scores, measured in standard units, is always right at the average, so in this case a height of $70$ inches would correspond to a $z$-score of $0$. A height of $74.5$ inches, being $4.5$ inches and therefore (as we saw before) $1.5$ standard units above average, would correspond to a $z$-score of $0+1.5=1.5$. And so on.

When you calculate $$z=\frac{x-\bar x}s$$ to get a $z$-score $z$ from a measurement $x$, you’re doing the same kind of unit conversion that you do in converting a temperature from one scale to the other. The calculation $x-\bar x$ gives you the deviation of your actual measurement from the mean; in my example, that’s $74.5-70=4.5$ inches. When you divide by $s$, the standard deviation, you’re changing ‘yardsticks’ from inches to standard units. In the example $s=3$ inches, so you’re multiplying the deviation of $4.5$ inches by the conversion (scaling) factor of $\frac13$ of a standard unit per inch.

One point of these standard units is that the permit comparison of distributions. For example, women are on average shorter than men, and their heights vary a bit less. Thus, a woman who is $3$ inches above the female average is, relative to the female population, taller than a man who is $3$ inches above the male average. But how much taller? Use of standard units makes it possible to answer that kind of question. I’m using data that are now a bit out of date, but very roughly she is $1.2$ standard units above the female average, while he is only $1$ standard unit above the male average.

  • $\begingroup$ .....excellent! $\endgroup$
    – Artegon
    Jan 11, 2013 at 15:38
  • $\begingroup$ @user1315357: Glad it helped. $\endgroup$ Jan 11, 2013 at 15:39
  • $\begingroup$ THANK YOU FOR THAT AWESOME EXPLANATION! So helpful - I swear I would be so stuffed if I had to just rely on the textbook and lectures for my stats course. Answers like that make all the difference. $\endgroup$
    – user142177
    Apr 10, 2014 at 8:03
  • $\begingroup$ "You might say that the standard deviation is a yardstick, and a z-score is a measurement expressed in terms of that yardstick." Fantastic. Thanks. $\endgroup$
    – jds
    Jan 27, 2016 at 19:37
  • $\begingroup$ @gwg: Glad it helped! $\endgroup$ Jan 27, 2016 at 19:54

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