# How do you estimate the normal distribution for large $z$?

In working out a recent blog post, I had cause to find the probability, in a standard normal distribution, of $z < -23$. Beyond "that's a REALLY SMALL NUMBER"1, I was stumped. Could I get anywhere close mentally? A good question.

Starting from the definition of the normal distribution, $p(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2}$, I figured that modelling the remainder of the graph as a triangle might be fruitful.

For $x = -23$, we're looking at the height of the graph, $p(-23) = \frac{1}{\sqrt{2\pi}} e^{-\frac{-529}{2}}$. I'm not going to work that out yet.

What's the gradient of the curve there? $\diff{p}{z} = \frac{x}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2}$, and at $z=-23$ that's $-23p(23)$. This is handy: it tells us that the height of our triangle, divided by its base, is 23 times the height -- so the base has to be $\frac{1}{23}$.

Let's recap: our height is $\frac{1}{\sqrt{2\pi}} e^{-\frac{-529}{2}}$ and our base is $\frac{1}{23}$, so our area is $P(z < 23) \approx \frac{1}{46\sqrt{2\pi}} e^{-\frac{-529}{2}}$.

And, in your head, that's... silly. Let's take logs instead: $\ln(P) \approx -264.5 - \ln(46 \sqrt{2\pi})$.

What's $\ln(46 \sqrt{2\pi})$? Well, $\sqrt{2\pi} \approx 2.5$ (it's slightly higher), so we want $\ln(115)$. Obviously, $\ln(128) = 7 \ln(2) \approx 4.9$ and $\ln(100) = 2(\ln(2) + \ln(5)) \approx 4.6$, so $\ln(115)$ is somewhere about 4.8.

OK! So $\ln(P) \approx -264.5 - 4.8 \approx -269.3$. But what's $e^{-269.3}$ as a proper number?

We can do a base conversion trick and work out $-\frac{269.3}{2.32}$ to get the log base 10. That's a bit bigger in magnitude than 100: doing it carefully gives about -116. We're out by a factor of 5, which is a pretty solid approximation, if you ask me!

In general, the form of this approximation is $\ln(P(x)) \approx -\frac{1}{2}x^2 - \ln(5|x|)$.

Trying it with $x=-7$, we get $\ln(P(7)) \approx -24.5 - \ln(35)$, and $\ln(35) \approx 3.5$, so $P(7)\approx e^{-28}$. Converting to base 10, $\frac{28}{2.32} \approx 12$ and a bit, so we're looking at a little more than $10^{-12}$. The correct answer is $1.4\times 10^{-12}$.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. Wolfram|Alpha says it's $2 \times 10^{-117}$ []

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