Inverse standard normal CDF
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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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 $0.841621$
$\endgroup$ 3 $\begingroup$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) ## 0.8416212Just 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 0.841621Approximations 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 tablefrom 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.)
This is not the exact answer to the question, but it is a solution that precludes the need for dealing with a lookup table. Also, relatively computationally inexpensive.
Equation 8 at gives a simple expression for the best logistic fit for the cumulative normal distribution:
$\phi(z) \approx \frac{1}{(1 + e^{-1.702z})}$
This is easily invertible as:
$z(\phi) \approx - \frac{ln (\frac{1}{\phi}-1)}{1.702}$
The accuracy of this expression is discussed at length in the referenced article. There are more accurate expressions there, but they're not quite so easy to invert and require more computational overhead. This one can be used for coding purposes where you want to generate "random" samples from a normal distribution
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