Assume that we have a set of *n* data samples from a Normal distribution with unknown mean *m* and unknown standard deviation *s*. We would like to estimate the mean together with the appropriate level of uncertainty. A Normal distribution can have a mean anywhere in [-∞, +∞], so we could use a Uniform improper prior *p**(m**)* = *k*. The uninformed prior for the standard deviation should be *p**(s**)* = 1/*s* to ensure invariance under a linear transformation. The likelihood function is given by the Normal distribution density function:

{i(X|\mu ,\sigma)=\frac{1}{(2\pi \sigma ^{2})^{n/2}}exp\Big(-\frac{1}{2 \sigma ^{2}}\displaystyle\sum_{i-1}^{n}(x_{i}-\mu)^{2}\Big)} |

Multiplying the priors together with the likelihood function and integrating over all possible values of *s*, we arrive at the posterior distribution for *m*:

{f(\mu)\propto[(n-1)\hat{\sigma} ^{2}+n(\bar{x}-\mu)^{2}]^{-n/2}} |

{f(\mu)\propto \Big[1+\frac{n(\bar{x} -\mu)^{2}}{(n-1)\hat{\sigma }^{2}}\Big]^{-n/2}} |

(1)

where x and are the mean and sample standard deviation of the data values. The Student-t distribution with *n* degrees of freedom has probability density:

{f(y)\propto\Big[1+\frac{y^{2}}{\nu}\Big]^{-\big(\frac{\nu+1}{2}\big)}} |

(2)

Equation 1 and 2 are the same functions if we set n = *n* – 1 and {y=\frac{\sqrt{n}(\bar{x}-\mu )}{\hat{\sigma}}}.

Thus:

{\mu =t(n-1)\frac{\hat{\sigma} }{\sqrt{n}}+\bar{x}} |

where *t(n-1)* represents the Student-t (0,1,n-1) distribution with (*n*-1) degrees of freedom. Therefore, with Crystal Ball 7.0+, we can make the following model as also shown in the model Estimate_mean_and_stdev_for_Normal_distribution_when_neither_known

*m* = Student (x, *s/*SQRT(n), n-1)

This is the same result used in classical statistics.