Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From a given z-score, what is the formula for calculating the probability of the deviation being due to chance?


Given z = 1.2 standard deviations

The chance probability p = 0.11507

How did we derive p? (i.e. the formula, without having to refer to the normal distribution table)

share|cite|improve this question
up vote 1 down vote accepted

The probability $p$ is given by $$p=\int_{1.2}^\infty \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\, dx.$$ Equivalently, it is $\dfrac{1}{2}$ minus the integral from $0$ to $1.2$ of the same thing.

The function we are integrating is the probability density function of the standard normal. Since $e^{-x^2/2}$ does not have an antiderivative that is expressible in closed form in terms of elementary functions, the definite integral has to be calculated in another way. The calculations are not simple, so an old-fashioned solution was to prepare a table.

share|cite|improve this answer
Informative answer, thank you. – Marven Dec 12 '12 at 21:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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