Confidence Interval

The probability that a measurement will fall within a given Closed Interval $[a,b]$. For a continuous distribution,

$$\mathop{\rm CI}\nolimits (a,b)\equiv\int_b^a P(x)\,dx,$$ (1)

where $P(x)$ is the Probability Distribution Function. Usually, the confidence interval of interest is symmetrically placed around the mean, so

$$\mathop{\rm CI}\nolimits (x)\equiv \mathop{\rm CI}\nolimits (\mu-x,\mu+x)=\int_{\mu-x}^{\mu+x} P(x)\,dx,$$ (2)

where $\mu$ is the Mean. For a Gaussian Distribution, the probability that a measurement falls within $n\sigma$ of the mean $\mu$ is

$$\mathop{\rm CI}\nolimits (n\sigma) \equiv {1\over\sigma\sqrt{2\pi}}\int_{\mu-n\sigma}^{\mu+n\sigma}e^{-(x-\mu)^2/2\sigma^2}\,dx$$

$$= {2\over\sigma\sqrt{2\pi}} \int_0^{\mu+n\sigma}e^{-(x-\mu)^2/2\sigma^2}\,dx.$$ (3)

Now let $u\equiv (x-\mu)/\sqrt{2}\,\sigma$, so $du=dx/\sqrt{2}\,\sigma$. Then

$$\mathop{\rm CI}\nolimits (n\sigma) = {2\over\sigma\sqrt{2\pi}}\sqrt{2}\,\sigma \int_0^{n/\sqrt{2}}e^{-u^2}\,du$$

$$= {2\over\sqrt{\pi}} \int_0^{n/\sqrt{2}} e^{-u^2}\,du = \mathop{\rm erf}\nolimits \left({n\over\sqrt{2}}\right),$$ (4)

where $\mathop{\rm erf}\nolimits (x)$ is the so-called Erf function. The variate value producing a confidence interval CI is often denoted $x_{\rm CI}$, so

$$x_{\rm CI} = \sqrt{2}\,\mathop{\rm erf}\nolimits ^{-1}(\mathop{\rm CI}\nolimits ).$$ (5)

range	CI
$\sigma$	0.6826895
$2\sigma$	0.9544997
$3\sigma$	0.9973002
$4\sigma$	0.9999366
$5\sigma$	0.9999994

To find the standard deviation range corresponding to a given confidence interval, solve (4) for $n$.

$$n=\sqrt{2}\,{\rm erf}^{-1}(\mathop{\rm CI}\nolimits )$$ (6)

CI	range
0.800	± 1.28155$\sigma$
0.900	± 1.64485$\sigma$
0.950	± 1.95996$\sigma$
0.990	± 2.57583$\sigma$
0.995	± 2.80703$\sigma$
0.999	± 3.29053$\sigma$