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In a standard normal distribution how do I deal with a $Z$ value greater than 3?

I know that z-score ranges form -3 to 3

Consider this one ...

mean = 70, standard deviation = 4

I need to find $P(65 < X < 85)$.

Transforming to standard normal gives $P(-1.25 < Z < 3.75)$

How to deal with the $3.75$?

Edit *Actually it's not greater than 3, $z < 3.75$. I meant smaller than, sorry. Should I assume it's just 0.4990 or what?

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    $\begingroup$ z scores range from $-\infty$ to $+\infty$. $\endgroup$
    – Patrick Coulombe
    Commented May 14, 2014 at 21:00
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    $\begingroup$ ok i know that but i'm talkin about the standard .. which has table with values from -3 from 3 , i know it's up to infinity but what i want is to find values bigger than 3 which is not listed in the table .. i know there is another exponential formula but it's not for the standard dist. $\endgroup$
    – Mr.Cat
    Commented May 14, 2014 at 21:03
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    $\begingroup$ The $p$-value of $|z|$ statistics that are $\ge 3$ are all small. They never get big again... just smaller and smaller. So if your table of $p$-values for $z$ statistics stops at three, it probably says the $p$-value is 0.000 or something like that (really meaning < 0.001), and you can simply use that to conduct your tests. $\endgroup$
    – Alexis
    Commented May 14, 2014 at 21:08
  • $\begingroup$ The 68-95-99.7 Rule already gives you an answer to better than two significant figures. $\endgroup$
    – whuber ♦
    Commented Dec 8, 2023 at 22:13

4 Answers 4

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Let me repeat (and correct) what I've said in my comment ad reply to your edit.

You have to transfer from $X$ to $Z$ in order to use a z-score table. Since a z-score table contains a small finite subset of values, you often must settle for an approximation. So you could also settle for $P(Z<-3)\approx 0$ and $P(Z< 3)\approx 1$ (NB: $P(Z>3)\approx 1$ was a typo, sorry.)

As to $P(-1.25<Z<3.75)$, I'll use this z-score table: $$P(-1.25<Z<3.75)=P(Z<3.75)-P(Z<-1.25)\approx 1-0.1056=0.8944$$

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The standard normal ranges from $-\infty$ to $\infty$.

Your problem appears to be that your table doesn't go further.

Your question should therefore be modified to ask "*How do I deal with the fact that my table doesn't go as high as my $Z$ value?*"

[Note that in your last paragraph, you have become confused. The region you're evaluating probability for is $Z<3.75$ but the boundary value of $Z$ you're trying to look up in the table (the $3.75$) is $>3$, as in your title.]

It seems like not having the value in your table would be a problem, but it's a very small one $-$ since your answer for $P(0<Z<3.75)$ can't be smaller than $P(0<Z<3)\approx 0.4999$ and can't be larger than $P(0<Z<\infty)=0.5$, you shouldn't have much difficulty narrowing the answer down to 3 significant figures of accuracy even so.

Additional accuracy (though I really don't think you need it) can be obtained by many methods. Here are three:

i) finding better tables (these seem to be of the same form as the ones you're apparently using)

ii) using a package that will evaluate standard normal cdfs for you. I just used R (simply typing pnorm(3.75) to obtain $P(-\infty<Z<3.75)$).

iii) using numerical integration to approximate the area between 3 and 3.75. For example, via Simpson's rule, a single interval (3 points) gives 0.0017 (the correct answer is 0.0013 to 4dp). Alternatively, because the density is convex in this region (indeed, as whuber points out in comments, convex for $Z>1$ and $Z< -1$), the integral will be bounded below by the midpoint rule and above by the trapezoidal rule, which usefully bounds where the answer can lie

But, really, just using the limits provided by 3 and $\infty$ is plenty, I imagine.

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  • $\begingroup$ +1. To head off a possible unintended reading of this answer, note that the density is not convex everywhere--only for $|Z|\gt 1$. $\endgroup$
    – whuber ♦
    Commented May 14, 2014 at 22:53
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    $\begingroup$ @whuber quite right - I intended to type "convex in this region" but somehow it didn't end up in the text. I will fix that, thanks. $\endgroup$
    – Glen_b
    Commented May 14, 2014 at 22:57
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For $z >0$, the right tail of the standard normal distribution (that is, the area to the right of $z$) which is often denoted by $Q(z)$ is bounded as follows:

$$\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z} - \frac{1}{z^3}\right ) \ < \ Q(z) \ < \ \frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right ).$$

See, for example this answer on math.SE for a proof. The bounds blow up/down to $\pm \infty$ as $z \to 0$ but are quite useful in the regions not covered by typical tables of the cumulative distribution function of the standard normal random variable. For example, $$0.0000873 < Q(3.75) < 0.0000940$$ while the actual value is slightly larger than $0.0000884$

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  • $\begingroup$ I pondered whether to include a calculation based on the series for Mill's ratio ($Q(z)/\phi(z)$) in my answer - I even typed in something under a point "iv)" but then deleted it instead of posting it. Glad to see it here. $\endgroup$
    – Glen_b
    Commented May 15, 2014 at 4:18
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    $\begingroup$ Should be one of Mills' ratio or Mills's ratio or Mills ratio: person concerned was John P. Mills whose name appeared on a tabulation he did for Karl Pearson. He didn't invent or discover the ratio and he published nothing else, it seems. $\endgroup$
    – Nick Cox
    Commented May 15, 2014 at 8:31
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    $\begingroup$ Connecting this answer with that of @Glen_b, it is interesting that a crude application of Simpson's Rule to the interval $[z, z+2h]$ with $h\approx 3/z$--beyond which the integrand clearly gets very small compared to its value at $z$--yields the approximation $\frac{1}{\sqrt{2\pi}}\exp(-z^2/2)/z$ with $O(z^{-3})$ error. $\endgroup$
    – whuber ♦
    Commented May 15, 2014 at 16:06
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    $\begingroup$ An even cruder approximation is as follows. For $z>0$, the tangent to $\phi(z)$, which has slope $-z\phi(z)$, crosses the horizontal axis at $z+z^{-1}$. For $z>1$, the right triangle whose hypotenuse is the tangent and whose base extends from $z$ to $z+z^{-1}$ lies wholly in the region that we integrate to find $Q(z)$. This triangle has area $\frac{\phi(z)}{2z}$ giving that $$Q(z) > \frac{1}{2}\cdot\frac{\exp(-z^2/2)}{\sqrt{2\pi}}\left (\frac{1}{z}\right )$$ which is half the upper bound. Crude, but better than the lower bound which has value $0$ at $z=1$ $\endgroup$
    – Dilip Sarwate
    Commented May 16, 2014 at 20:31
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The z (i.e., normal) distribution is not bounded. $\mathcal N(\mu=70,\sigma=4)$ is not standard normal either – that refers to $\mathcal N(0,1)$. If you're wondering what the p value is for z = 3.75, you can find it in with pnorm(3.75). (You could also use pnorm(85,70,4).) The result is p = 0.9999116.

If you want an exact p value, I think you're going to have an easier time getting it in R or some other statistical software than by dealing with the quantile function directly...but FWIW, here's that equation:

$$F^{-1}(p) = \mu + \sigma\Phi^{-1}(p) = \mu + \sigma\sqrt2\,\operatorname{erf}^{-1}(2p - 1), \quad p\in(0,1)$$

In the above, $\rm erf$ refers to the error function.

In light of your comments and edit to the question, I think I should decline to provide more than this as a "hint" in following our policy on self-study questions.

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  • $\begingroup$ i think i'v been learning things the wrong way ! i didn't got your answer, but in class they use this formula to transfer from X to Z Z = (X-μ)/σ ---> (X-MEAN)/STD am i right ? $\endgroup$
    – Mr.Cat
    Commented May 14, 2014 at 20:54
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    $\begingroup$ Basically, yes, that's the way to z-score. FWIW, using a sample mean and standard deviation in place of the population parameters yields t-scores instead. I'm not sure I understood what you were after, but there is nothing particularly difficult about handling z scores larger than 3, except maybe not being able to look up corresponding p values in a textbook table. IMO, it's better to leave the table behind and start calculating statistics anyway. $\endgroup$
    – Nick Stauner
    Commented May 14, 2014 at 21:02
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    $\begingroup$ @Mr.Cat, you have to transfer from $X$ to $Z$ in order to use a z-score table. Since a z-score table contains a small finite subset of values, you often must settle for an approximation. So you could also settle for $P(Z<3)\approx 0$ and $P(Z>3)\approx 1$. $\endgroup$
    – Sergio
    Commented May 14, 2014 at 21:02
  • $\begingroup$ @Sergio yes i've been told something like this related to approximation ! $\endgroup$
    – Mr.Cat
    Commented May 14, 2014 at 21:10
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    $\begingroup$ @Sergio your inequalities contain typographical errors (they both go the wrong way). No doubt you intended them the other way around, but I think I have to mention it so that later readers are aware that $P(Z<3)\approx 1$ and $P(Z>3)\approx 0$ was what you meant. $\endgroup$
    – Glen_b
    Commented May 17, 2014 at 2:19

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