Why $z(0.995)$ is $2.58$ and not $2.575$ Why some textbooks say that z(0.9950)=2.58, for instance "Statistics" by Murray R. Spiegel.
Why don't they interpolate? If you look up in the z-table z(0.9949)= 2.57 and z(0.9951)=2.58
Thanks for all the answers. I understand that z(0.9950) is closer to 2,58 than to 2.57. But then, why the same book states that z(0.95)=1.645? Why don't they round off the number to 2 decimal digits? (1.640 or 1.650).
Where can I find a z-table with more than 4 decimal digits?
And a a z-table for z>4 or greater ?
Thank you for the answers and the links.
But what is the reason of using 
2 decimal digits for  z(0.9950)=2.58
and 3 decimal digits for  z(0.9500)=1.645
(both numbers taken from the same table, talking about confidence levels)
 A: Here's one reason for reporting different numbers of digits for $z(.95)$ versus $z(.995)$: if you look at the graph of the inverse of the cumulative distribution function of the normal distribution, if you plot the two points $(.95,z(.95))$ and $(.995,z(.995))$ on that graph, and if you look at the slopes of the graph at those two points, you will find that the slopes are quite different.
For instance, in the appendix of my textbook, the table for the c.d.f. $\Phi(z)$ gives the following values:


*

*Near the point $(.95,z(.95))$ we have the point $(.9495,1.64)$ given by $.9495=\Phi(1.64)$, and the point $(.9505,1.65)$ given by $.9505=\Phi(1.65)$, and so the slope of graph of the inverse c.d.f. is about $\frac{\Delta z}{\Delta \Phi} = \frac{.01}{.9505-.9495} = 10$

*Near the point $(.995,z(.995))$ we have the point $(.9949,2.57)$ given by $.9949=\Phi(2.57)$, and the point $(.9951,2.58)$ given by $.9951=\Phi(2.58)$, and so the slope of the graph of the inverse c.d.f. is about $\frac{\Delta z}{\Delta \Phi} = \frac{.01}{.9951-.9949} = 50$


The key point is that the slope is around 5 times as steep near the input $.995$ as it is near the input $.95$. Thus, the effect on $z$ arising from small changes near $.995$ and the effect on $z$ from identical small changes near $.95$ are off by a rather large 5-to-1 ratio. So it makes sense to report 1 fewer digit of $z$ near $.995$ than near $.95$, because if we do so then the effect on the least significant digit of $z$ reported near $.995$ and the effect on the least significant digit of $z$ reported near $.95$ are off by a somewhat smaller 1-to-2 ratio.
A: The standard normal cumulative distribution function is 
$$
F(x) = \frac{1}{2} \left(  1 +  \mathrm{erf}(x/\sqrt{2} ) \right) 
$$
where "erf" is the error function.   Solving the equation $F(x)=0.995$  (or any probablility) other to arbitrary precision is possible, but not terribly useful as stated in an earlier comment.  But if you a more accurate number, here is the result to 50 digits: 
$$
x=2.5758293035489007609785767486038141173060176342763
$$
