1
$\begingroup$

We are dealing with a standard normal random variable. We have $$\Phi(c) = 0.8$$ where $c$ is just some arbitrary number and $\Phi$ is just the usual notation for the CDF of a standard normal distribution.

I want to find such a $c$ so that this equation holds, i.e.: $$c = \Phi^{-1}(0.8)$$ where we just take the inverse function. How do I find this on a z-table?

$\endgroup$
0
$\begingroup$

It depends, if you have a table that gives "area to the left", then look up the row that has area = .8. The standardized score $c$ should be $0.8416212$.

Some tables are defined as area in the middle, then that's a little trickier.

$\endgroup$
1
$\begingroup$

This is not an answer to the question since it gives approximate formulae for the calculation to replace table lookup. Moreover, I am not a statistician : so, more than likely, I am discovering the wheel !

If we are not concerned by the ends of the curve, we can express the inverse of the cumulative normal distribution function using Padé approximants $P_{m,n}$ built around $p=\frac 12$ .

Being lazy, I just focused on $m=n$ and I give you below some formulae you could play with $$P_{2,2}=\frac{\sqrt{2 \pi } \left(p-\frac{1}{2}\right)}{1-\frac{1}{3} \pi \left(p-\frac{1}{2}\right)^2}$$ $$P_{3,3}=\frac{\sqrt{2 \pi } \left(p-\frac{1}{2}\right)-\frac{11 \pi ^{3/2} }{15 \sqrt{2}}\left(p-\frac{1}{2}\right)^3}{1-\frac{7}{10} \pi \left(p-\frac{1}{2}\right)^2}$$ $$P_{4,4}=\frac{\sqrt{2 \pi } \left(p-\frac{1}{2}\right)-\frac{157}{231} \sqrt{2} \pi ^{3/2} \left(p-\frac{1}{2}\right)^3}{1-\frac{78}{77} \pi \left(p-\frac{1}{2}\right)^2+\frac{241 \pi ^2 }{2310}\left(p-\frac{1}{2}\right)^4}$$ For the example $p=0.8$, they respectively give $0.830236$, $0.84035$, $0.841473$ for an exact value equal to $0.841621$

$\endgroup$
  • $\begingroup$ "How do I find this on a z-table?" $\endgroup$ – symplectomorphic Mar 9 '16 at 8:07
  • $\begingroup$ Lots of activity lately in approximations to std normal CDF, especially if invertible. One of many is http//lyle.smu.edu/~aleskovs/emis/sqc2/accuratecumnorm.pdf. I understand R now uses an invertible approx for its rnorm function--instead of Box-Muller. $\endgroup$ – BruceET Mar 14 '16 at 21:14
  • $\begingroup$ @BruceET. Thank you for this comment. This is an area (one more) where my knowlegde is close to $\epsilon$; I just tried something simple. Thnaks for the link. By the way, I am always impressed by the quality of your answers on MSE. Cheers. $\endgroup$ – Claude Leibovici Mar 15 '16 at 3:12
0
$\begingroup$

The inverse of the CDF is often called the quantile function.

Software solutions. If you are using software there is usually a way to get quantile functions. For example in R, the inverse of $\Phi$ is qnorm (with default $\mu = 0$ and $\sigma = 1$), and in Minitab it's the command INVCdf followed by subcommand NORM 0 1. Other software packages have their own syntax.

In R, the answer to your specific question would be obtained as follows:

 qnorm(.8)
 ## 0.8416212

Just to check on this, the R code for the standard normal CDF is pnorm, and the statement pnorm(0.8416212) returns 0.8 exactly.

Here's how it looks in Minitab:

 MTB > invcdf .8;
 SUBC> norm 0 1.

 Inverse Cumulative Distribution Function 

 Normal with mean = 0 and standard deviation = 1

 P( X <= x )         x
         0.8  0.841621

Approximations from printed tables. However, you asked about printed tables. Suppose it's a straightforward CDF table. Then you look around in the $body$ of the table to find the entry nearest to .8. In the table I'm looking at, I find the entry .7995 corresponding to z-value .84 in the $margins$ of the table. So without interpolation .84 is as close as I can get. Linear interpolation between entries .7995 and .8023 (corresponding to $z = .85$) would get me a little closer: $.84 + 0.01(5/28) = 0.8418,$ which is wrong in the fourth place. But that's about as accurate an answer as you'll get from a printed table.

Notes: (1) I know of printed tables that give probabilities to five places rather than four, but all the ones I have seen give z-values only to two places. (Perhaps see the five-place table from the NIST online handbook.) (2) As software is ever more widely used, I suppose printed tables will disappear at some point. I know of a few recent probability and statistics books that have no normal tables. So maybe I'm not totally wasting your time by showing some software answers. (3) Not to be too fussy, but the usual notation for the standard normal CDF is $\Phi$, not $\phi$. In TeX, you get it by typing $\Phi$, instead of $\phi$. Sometimes $\phi$ is used for the standard normal PDF. (I have edited your Question accordingly.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.