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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,
\end{displaymath} (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,
\end{displaymath} (2)

where $\mu$ is the Mean. For a Gaussian Distribution, the probability that a measurement falls within $n\sigma$ of the mean $\mu$ is
$\displaystyle \mathop{\rm CI}\nolimits (n\sigma)$ $\textstyle \equiv$ $\displaystyle {1\over\sigma\sqrt{2\pi}}\int_{\mu-n\sigma}^{\mu+n\sigma}e^{-(x-\mu)^2/2\sigma^2}\,dx$  
  $\textstyle =$ $\displaystyle {2\over\sigma\sqrt{2\pi}} \int_0^{\mu+n\sigma}e^{-(x-\mu)^2/2\sigma^2}\,dx.$ (3)

Now let $u\equiv (x-\mu)/\sqrt{2}\,\sigma$, so $du=dx/\sqrt{2}\,\sigma$. Then
$\displaystyle \mathop{\rm CI}\nolimits (n\sigma)$ $\textstyle =$ $\displaystyle {2\over\sigma\sqrt{2\pi}}\sqrt{2}\,\sigma \int_0^{n/\sqrt{2}}e^{-u^2}\,du$  
  $\textstyle =$ $\displaystyle {2\over\sqrt{\pi}} \int_0^{n/\sqrt{2}} e^{-u^2}\,du = \mathop{\rm erf}\nolimits \left({n\over\sqrt{2}}\right),$ (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 ).
\end{displaymath} (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 )
\end{displaymath} (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$\sigma$

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© 1996-9 Eric W. Weisstein