Can you have a z score that extends past -5 or +5?

Do z-scores run from negative infinity to positive infinity?

• If the support of the distribution is wide enough and the tails are thick, this is certainly possible. – Dimitriy V. Masterov Jun 17 '16 at 1:02
• If you are thinking of a particular data set it might be useful to provide details, in case the real question here is 'are extreme z-scores plausible in this particular instance'. – Robert de Graaf Jun 17 '16 at 1:54

You can certainly get a z-score to exceed 5 in absolute size, or indeed any other finite value.

However, if you're internally standardizing (i.e. using the sample mean and sample standard deviation of the same sample that you're standardizing) to calculate your z-scores then there is a bound on the size of the largest absolute z-score at any given sample size.

So while z-scores can theoretically exceed any bound, when standardizing by the sample statistics, you may not be able to do it at some given sample size.

For example at $n=26$ you can't quite reach a z-score of 5 but at $n=27$ you can exceed it.

Here's an example data set with n=27 that has a z-score whose absolute value exceeds 5:

78.7  79.2  79.3  79.3  79.5  79.6  79.6  79.8  79.8  79.9  80.2  80.2
80.2  80.3  80.5  80.6  80.8  81.0  81.1  81.1  81.4  81.6  81.7  81.7
81.7  82.0 600.0


internally standardized z-scores:

-0.210 -0.205 -0.204 -0.204 -0.202 -0.201 -0.201 -0.199 -0.199 -0.198
-0.195 -0.195 -0.195 -0.194 -0.192 -0.191 -0.189 -0.187 -0.186 -0.186
-0.183 -0.181 -0.180 -0.180 -0.180 -0.177  5.003


(Note that if you remove any data point but the last, the z-score of that "600" drops below 5 (to about 4.903). This example doesn't prove it can't be done at $n=26$, of course, but it can't in any case.)

More generally, if you place $n^2+1$ values at $\frac{-1}{\sqrt{n^2+2}}$ and one value at $\frac{n^2+1}{\sqrt{n^2+2}}$ then the mean will be $0$ and the standard deviation will be $1$ (i.e. these will be internally standardized z-scores), and so we have a way to make the largest z-score in absolute size (by the above construction) to be $\frac{n^2+1}{\sqrt{n^2+2}}>n$.

Yes and yes. For example, if you transform a vector with a hundred 1s and a single 2 to z-scores, the 2 will become a z-score of 10. Replace the 2 with -1 and it'll get transformed to -10 instead.

By $z$-scores we mean subtracting the mean and dividing by standard deviation. So to obtain $z$-score of five or more you need $|x-\bar x|$ simply to be greater than five standard deviations...

Moreover, from Chebyshev's inequality we know that no mather what is the distribution of your data

$$\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}$$

so probability of observing $x$'s distant by five standard deviations from the mean is at most $1/25$. Even if you were thinking of normal distribution and the "three sigma" rule, then $z$-scores of $-5$ or $5$ would appear in less than $1\%$ cases buth with non-zero probability.

In theory if $x \in (-\infty, \infty)$, then obviously $z = \frac{x-\mu}{\sigma}$ can be infinite. However in real life you cannot observe inifinite values, so $z$-scores cannot be infinite. They can lead to huge values, but not infinite.