# z score of normal distribution

Good day,

I want to ask about standard normal distribution. What is the highest and lowest value of $z$ score can be?

From the table of standard normal, the value $z$ score is only for -3.99 $\leq$ $z$ $\leq$ 3.99.

Can it be higher than the range? Is there any programme or formula to compute the probability of standard normal distribution for the range outside -3.99 $\leq$ $z$ $\leq$ 3.99?

The table below shows the values given by R for $\Phi(x)=\Pr(X \le x)$ for some $x$. In addition $\Phi(-x)=-\Phi(x)$.

For $x$ large and negative, a reasonable approximation is $\dfrac{-x}{x^2+1}\dfrac{\exp\left(-x^2/2\right)}{\sqrt{2\pi}}$. For example, with $x=-15$, this gives 3.670825e-51

So for $x$ large and positive, a reasonable approximation is $1-\dfrac{x}{x^2+1}\dfrac{\exp\left(-x^2/2\right)}{\sqrt{2\pi}}$.

   x  Phi(x) =  Pr(X<=x)
-15    3.670966e-51
-14    7.793537e-45
-13    6.117164e-39
-12    1.776482e-33
-11    1.910660e-28
-10    7.619853e-24
-9    1.128588e-19
-8    6.220961e-16
-7    1.279813e-12
-6    9.865876e-10
-5    2.866516e-07
-4    3.167124e-05
-3    1.349898e-03
-2    2.275013e-02
-1    1.586553e-01
0    5.000000e-01
1    8.413447e-01
2    9.772499e-01
3    9.986501e-01
4    9.999683e-01
5    9.999997e-01

• How to get these values? From books or programme? Commented Oct 9, 2014 at 7:22
• From R's pnorm function. Commented Oct 9, 2014 at 8:38
• Thank you for your explanations. But I guess the formula is only approximation for cumulative normal distribution. Because if I let x=-1 into the formula, i couldn't get the same value as in table of normal distribution. Am I right? Commented Oct 9, 2014 at 9:36
• @xzatulx: $-1$ is not large and you have its value in your tables. The approximations were suggestions for $(-\infty,-4]$ and $[4,\infty)$ Commented Oct 9, 2014 at 10:54

$z$ score that is above $3.999$ or below $-3.999$ is considered highly unusual in Triola book which means that it very rarely occurs. Any other value of $z$ that is greater than $3.999$ is treated the same as $3.999$ or $-3.999$ in the case of negative $z$ because $P(|z| > 3.999) \approx 0$

• Thank you for the response. Can the probability be exactly value of zero or 1? Commented Oct 9, 2014 at 7:20