Finding normal probabilities

Finding P(X < a), P(X > a) and P(a < X < b) for a normal distribution as areas under the curve, and the inverse problem — finding the value x for a given probability (the inverse normal).

Statistics · Statistical distributions

Finding normal probabilities

Finding P(X < a), P(X > a) and P(a < X < b) for a normal distribution as areas under the curve, and the inverse problem — finding the value x for a given probability (the inverse normal).

Why it works

A probability for a normal variable is an area under the bell curve. Modern calculators give these areas directly from

\mu

and

\sigma

, but the thinking is always about which area you want.

$P(X < a)$ is the area to the left of $a$ (the shaded region above, with the

curve standardised).

$P(X > a) = 1 - P(X < a)$ — the area to the right is the complement.
$P(a < X < b) = P(X < b) - P(X < a)$ — subtract the smaller left-area from the

larger.

If you standardise by hand,

P(X < a) = P\!\left(Z < \dfrac{a - \mu}{\sigma}\right) = \Phi(z)

, where

\Phi

is the standard normal area-to-the-left. The symmetry

\Phi(-z) = 1 - \Phi(z)

handles negative

z

.

The inverse problem. Sometimes you're given the probability and must find the value — "the mark exceeded by the top

10\%

", say. This is the inverse normal: find

z

with

\Phi(z) = p

(the inverse normal function, or table), then convert back with

x = \mu + z\sigma.

For "the top

10\%

P(X > x) = 0.10

means

P(X < x) = 0.90

, so

z

is the value with

\Phi(z) = 0.90

, namely

z \approx 1.2816

, and

x = \mu + 1.2816\,\sigma

.

A sketch of the bell with the wanted area shaded prevents nearly every mistake — it shows instantly whether you need a left area, a right area, or its complement.