# A standard normal integral calculator

The little tool below allows you to easily compute standard integrals, which have the form $$P(Z \lt z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^z e^{-x^2/2}dx.$$

To compute $P(Z \lt z)$, simply enter $z$ into the box below.

Note that the integrand is exactly the standard normal distribution so, if $Z$ is a standard, normal random variable and $z$ is a fixed number, then the integral represents the probability that $Z$ is less than $z$ - which is exactly what we mean by $P(Z \lt z)$. The number $z$ is often called the Z-score.
More generally, we say that $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ if its distribution function is $$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma^2)}.$$ The standard normal is exactly the normal with $\mu=0$ and $\sigma=1$. While the normal calculator here only computes integrals for the standard normal, we translate any normal integral to an integral involving the standard normal using the $u$-substitution $$u = \frac{x-\mu}{\sigma}.$$ For then $\displaystyle du = \frac{1}{\sigma}\,dx$ and $$\frac{1}{\sqrt{2\pi}\sigma}\int_a^b e^{-(x-\mu)^2/(2\sigma^2)} dx = \frac{1}{\sqrt{2\pi}}\int_a^b e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\,\frac{1}{\sigma}\,dx = \frac{1}{\sqrt{2\pi}} \int_{(a-\mu)/\sigma}^{(b-\mu)/\sigma} e^{-u^2/2}du.$$ In statistics, this is called computing the Z-score.
Suppose that $X$ is normall distributed with mean $\mu=74$ and standard deviation $\sigma = 10$. Compute $P(70 \lt X \lt 80)$.
Solution: Expressed as normal integral using the mean and standard deviation of $X$, we have $$P(70 \lt X \lt 80) = \frac{1}{\sqrt{2\pi} \times 10} \int_{70}^{80} e^{-(x-74)^2/(2\times10^2)}dx.$$ Letting $\displaystyle u = \frac{x-74}{10}$, this can be re-written as $$\frac{1}{\sqrt{2\pi} \times 10} \int_{70}^{80} e^{-(x-74)^2/(2\times10^2)}dx = \frac{1}{\sqrt{2\pi}} \int_{70}^{80} e^{-\frac{1}{2}\left(\frac{x-74}{10}\right)^2}\frac{1}{10}dx = \frac{1}{\sqrt{2\pi}} \int_{-0.4}^{0.6} e^{-u^2/2}du.$$ To compute this integral, we can plug $0.6$ and $0.4$ into the calculator and take the difference. That works since $$\frac{1}{\sqrt{2\pi}} \int_{-0.4}^{0.6} e^{-u^2/2}du = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0.6} e^{-u^2/2}du - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{-0.4} e^{-u^2/2}du.$$ Carrying that out, we get $$0.725747 - 0.344578 = 0.381169.$$