12
$\begingroup$

It's my understanding that a z-score can only be calculated and accurately used for data sets that are normally distributed.

Since "perfect" normal distribution almost never occurs in real-world data (where "perfect" normal distribution is defined as 1. The mean, median, and mode all equal the same number, 2. the distribution is perfectly symmetrical between all standard deviations on both sides of the mean, and 3. the distribution is asymptotic), how "close" can the distribution be to perfectly normal for the z-score to still be a valid statistical measure?

If the answer is it has to be a perfectly normal distribution, then my question is why is the z-score so important since it would seldom be a valid measure on real-world data?

$\endgroup$
3
  • 1
    $\begingroup$ Could you explain what you mean by "accurately used" and "valid measure"? $\endgroup$
    – whuber
    Commented Nov 17, 2014 at 1:19
  • 1
    $\begingroup$ What you describe as "normal distribution" (the "perfect" is not needed -instead we attach "approximate" when all these properties do not hold exactly), is the description of any unimodal, symmetrical around its mean distribution, with support over the extended reals: Laplace, Logistic, and t-distribution come easily to mind, alongside the normal. $\endgroup$ Commented Nov 17, 2014 at 2:30
  • 2
    $\begingroup$ It can be calculated for anything you can compute means and standard deviations for. No matter what the distribution, it measures the number of standard deviations each observation is from the mean. As for whether it's all that meaningful when the distribution isn't reasonably close to normal, that's another question - one that would partly depend on the application. [It's a tool I use extremely rarely, but in most of my applications the measurements I generally deal with are meaningful quantities without standardizing.] $\endgroup$
    – Glen_b
    Commented Nov 17, 2014 at 7:22

4 Answers 4

16
$\begingroup$

Technically, z-scoring does not depend on any distributional assumptions, such as normality. It's just a way of describing how far observations are from the mean, no matter what the distribution happens to be. So no harm in z-scoring non-normal variables.

The main caveat is that z-scores tend to be more informative for distributions that are at least approximately symmetric about the mean (which includes normal distributions, but also many others), and less informative for highly skewed distributions. The reason is because in skewed distributions, two observations that lie on opposite side of the mean can have the same absolute z-score despite one being much more/less probable than the other.

$\endgroup$
1
  • $\begingroup$ Best and simplest explanation of this I've read so far. Thanks! $\endgroup$ Commented Sep 1, 2022 at 23:31
9
$\begingroup$

Not exactly, z-score is just a form a transformation. What it does is replacing the measurement unit with "number of standard deviations" away from the mean. Hence, it's a convenient tool when someone wants to compare two variables that are measured in different units.

What you mentioned about normal distribution is not really a condition for using z-score, but an additional perk to z-score interpretation. When a distribution is normal, approximately 68% of the data will be between -1 to 1 SD, 95% between -2 to 2 SD, and 99.7% between -3 to 3 SD. (see image here). The same, however, cannot be said for non-normal distributions.

$\endgroup$
4
  • 1
    $\begingroup$ Ah, super. Thank you!! I think my mistake is that you just can't use the z-score table if this distribution is not normal. Does this make sense? $\endgroup$
    – Gary
    Commented Nov 17, 2014 at 1:46
  • $\begingroup$ @Gary Correct, for instance if the distribution is not normal then the area at the right end may no longer be 2.5% at z-score of 1.96. $\endgroup$ Commented Nov 17, 2014 at 1:51
  • $\begingroup$ Penguin_Knight: Terrific, thanks so much for the quick reply. You've been a HUGE HELP!! $\endgroup$
    – Gary
    Commented Nov 17, 2014 at 1:54
  • $\begingroup$ I know that we transform data into z-score when two measures are in different units. If I want to calculate amount of calories lost per mins but in my data calories data range from 0~10 and minute range from 0 ~ 10000, then do I transform both of them to z-score then divide calorie/min? $\endgroup$
    – haneulkim
    Commented Nov 15, 2019 at 1:54
1
$\begingroup$
Z Score Formula = (x – μ) / ơ

There is no relation distribution with z-score. To find out z score you just need three things. data point(x), mean(μ) and standard deviation(ơ). It does not limit to any specific distribution. To transform any data point to a z-score is called standardization and z-score has no unit.

$\endgroup$
3
  • 1
    $\begingroup$ Fine, but what does this add to previous answers? $\endgroup$
    – Nick Cox
    Commented Oct 31, 2019 at 13:28
  • $\begingroup$ z-score has no unit. $\endgroup$ Commented Nov 1, 2019 at 12:26
  • $\begingroup$ Indeed; that is tacit in other answers but not explicit. So, why not improve your answer to explain why it has no units? $\endgroup$
    – Nick Cox
    Commented Nov 1, 2019 at 12:41
-2
$\begingroup$

I still makes sense to approximate the data as normal in certain instances...for most instances you would need to use substitute non-parametric tests such is Mann-Whitney etc.Also statisticians substitute Z scores with T scores to get more valid estimation read more about Z tables and z score...

$\endgroup$
1
  • 1
    $\begingroup$ The terseness and vagueness of this post (with its generic references to "certain instances" and "most instances") makes it all too possible for readers to interpret it in ways that are seriously mistaken. The last statement about substituting is surprising; I cannot find any way to interpret it that would make it correct. Perhaps you could offer more explanation? $\endgroup$
    – whuber
    Commented Apr 17, 2015 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.