How can I deal with sample distribution table when Z is greater or equal to 5?

For example:(-2.04 <z<-5.96) Sample distribution table value for -2.04 will be 0.2018 then what will be the sample distribution table value for -5.96?

  • 4
    $\begingroup$ $-2.04 <z<-5.96$ does not make sense, since $-2.04 >-5.96$ $\endgroup$
    – Henry
    Commented Jun 4, 2022 at 18:55
  • 2
    $\begingroup$ There exist good, practicable formulas for extreme values: see stats.stackexchange.com/questions/7200. However, the only time this will matter is when you are concerned with events that consist only of extreme values, such as $|Z|\gt 8.$ (An "extreme" value would be one beyond the range of arguments in your table.) $\endgroup$
    – whuber
    Commented Jun 5, 2022 at 14:03

1 Answer 1


According to @Dave's Comment, I think you want to evaluate $P(-5.96 < Z < -2.04)$ for standard normal random variable $Z.$

To start, that would be $P(Z < -2.04) - P(Z < -5.96).$ Various styles of CDF tables of the standard normal distribution are in common use, so it is difficult to know exactly what table you have.

The following figure illustrates the steps you need to take. You want the area under the normal curve (blue) between the two vertical dotted lines.

enter image description here

First, find $P(Z < -2.04) = 0.0207.$ If your table has negative values of $z,$ you can look in the margins of the table to find -2 (vertical headers) and 0.04 (horizontal headers). The corresponding number in the body of the table will be 0.0207 (rounded to four places).

If your table has only positive values of $z,$ then look for $P(Z > 2.04) = 0.0207 = P(Z < -2.04).$ because of the symmetry of the density function about $0.$

The area under the blue density curve to the left of the dotted red line is $0.0207.$

Next, you need to find $P(Z < -5.96) \approx 0.$ You can see that this is true, just looking at the figure. Also, notice that the table ends short of $-6$ or $6,$ precisely because there is essentially $0$ probability so far out into the tails of a standard normal distribution.

The area under the blue density curve to the left of the dotted brown line is very nearly $0.$

Thus $P(-5.96 < Z < -2.04) = 0.0207,$ correct to four places.

Note: If you are using a statistical software program there will be a way to find this probability directly. Also, some statistical calculators have the same capability.

In R statistical software the following statement finds $P(-5.96 < Z < -2.04)$ to many decimal places, and you can round the result to four places:

diff(pnorm(c(-5.96,-2.040), 0, 1))
[1] 0.02067516

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