In standard normal distribution table how I deal with value of Z greater than or equal to 5 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?

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.

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

