I read about the 68-95-99.7 rule that shows

probability levels

I also read that a Z value of 1.96 gives a confidence level of 95%

Is it correct to think of Z as the "number of standard deviations" ? Only I have never seen it mentioned that way.

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    $\begingroup$ For future readers of this page. Don't confuse the statistical ratio z (which this question refers to) with the Z-factor used to assess the signal-to-noise ratio of a screening assay. They are not at all related. graphpad.com/support/faq/… $\endgroup$ Jul 29, 2021 at 20:02

3 Answers 3


No. The z score is not 'the number of standard deviations'. Instead the z-score of a value is the number of standard deviations that value is above the mean. A z-score of 1.7 is 1.7 standard deviations above the mean. A z score of -1 is one standard deviation below the mean, and so on.

This is not mere nitpicking, it's essential to correctly conveying your meaning. I have seen exactly this imprecision in relation to z-scores lead to error on numerous occasion. Stats is not the place for woolly thinking and muddled words $-$ it is tricky enough when you say exactly what you mean.

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    $\begingroup$ Yes, thanks, fixed; my phone's not a great way to write answers and I do sometimes miss the typos (or in many cases, stray autocorrects). You're also free to edit answers when there's an annoying error as long as it doesn't alter the intent. $\endgroup$
    – Glen_b
    Jul 30, 2021 at 6:59
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    $\begingroup$ An edit needs 6 characters, so I didn't know what other 5 to put in ;-) $\endgroup$
    – Kirsten
    Jul 30, 2021 at 7:08
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    $\begingroup$ I know the answer claims it is not mere nitpicking, but it really seems like nitpicking to me. A z score is "not the number of standard deviations", but it is "the number of standard deviations ..." $\endgroup$
    – Curt F.
    Jul 30, 2021 at 15:41
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    $\begingroup$ @CurtF. Suppose the mean is 10, the SD is 2, and our score is 14. Interpreting z-score as "the number of standard deviations" leads to calculating the z-score as 14/2 = 7, whereas it should be (14-10)/2 = 2. For somebody familiar with stats that over-literal interpretation is so obviously wrong that we might not even consider it (I didn't until I read Glen's answer) but it's plausible that a novice might take it too literally. $\endgroup$ Jul 31, 2021 at 0:24
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    $\begingroup$ @GeoffreyBrent Thanks for the explanation, it actually didn't occur to me even after reading Glen_b's answer. Spelling it out could improve the answer. $\endgroup$
    – Gala
    Aug 1, 2021 at 8:39

Yes. A Z value of a particular data point tells you how many standard deviations it is from its mean. Z=0 means it has the same value as the population mean, Z=-1 means it is 1std lower than its mean etc. The probability that an observation will lie within the interval of its population mean plus/minus two times the standard deviation is 95%. This is the connection between z scores and confidence intervals.

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    $\begingroup$ "... given that it is sampled from a normal distribution" is not right. You can standardize any data with a defined mean and standard deviation, turning it into Z-scores. By definition, the Z-score represents number of standard deviations away form the mean, regardless of the distribution. This generally isn't terribly useful for non-normal distributions, but a Z-score of 1 always means that the value is 1 standard deviation above the mean, no matter what the distribution is. $\endgroup$ Jul 29, 2021 at 19:35
  • $\begingroup$ I edited my answer conform your suggestion, but now I am uncertain again. I agree that you can standardize any data with defined mean and standard deviation regardless of distribution. But this transformation does not necessarily result in a bell shaped curve. I included the sampling condition, because the questionner already mentioned the 68-95-99 rule. For this rule to have meaning the original dataset should be sampled from something approximately normal, right? $\endgroup$ Jul 30, 2021 at 20:37
  • $\begingroup$ @confusedstudent The 68-95-99 rule requires normalcy, but Z-scores can be useful even when we can't apply that rule. For instance, Chebychev's inequality](en.wikipedia.org/wiki/Chebyshev%27s_inequality) gives us a 0-75-89-94-... bound even for non-normal distributions. (That is, at least 3/4 of observations must have a Z-score of magnitude <= 2, 8/9 <= 3, 15/16 <= 4, and so on.) This is considerably looser than the 68-95-99 rule, but it can still be useful as a bound on rare events. $\endgroup$ Jul 31, 2021 at 0:35
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    $\begingroup$ @confusedstudent The 68-95-99 rule only applies to normally distributed variables. You've removed the normality requirement correctly in the Z-score standard deviation definition, but need to put it back in for the probability statement - "The probability that an observation will lie within the interval of its population mean plus/minus two times the standard deviation is 95%, for normally distributed variables." For non-normal distributions, Z=1 is still one standard deviation from the mean, but 68% of values won't be between Z=1 and Z=-1. $\endgroup$ Aug 2, 2021 at 17:10


Sometimes z-score refers to a quantile randomized z-score, where the quantiles of distribution are mapped to z of a standard normal, so by construction, z-score of one is bigger than 68.27% percent of values in the distribution, regardless of how many standard deviations from a mean a value is.

  • $\begingroup$ wouldn't it be exactly? not bigger? $\endgroup$
    – Kirsten
    Jul 30, 2021 at 20:20
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    $\begingroup$ @KirstenGreed I suppose I was a bit imprecise in my language. What I meant is that if you score for example on an IQ test in a 68th percintile, that would mean your score is bigger than 68% of people who took the test, and your IQ z-score would be 1 $\endgroup$
    – rep_ho
    Jul 30, 2021 at 20:32

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