The Normal Distribution N( $/mu ,/sigma ^{2}$)

The values of many quantitative variables are concentrated around a certain value and larger deviation are rather rare. This concentration (better density) could be represented by a function as shown in Figure 2.7. Traditionally the normal distribution is used as approximation of such a behavior. Under certain assumptions it is also possible to show mathematically that the distribution of a sum of many independent small errors tends to the normal distribution.

The density depends on 2 parameters $/mu ,/sigma ^{2}$ and is defined by

where

The location parameter $/mu$ corresponds to the expectation (mean) and the scale parameter $/sigma$ is called deviation or standard deviation (see Figure 2.7).

Figure 2.7: Density of the Normal Distribution with Mean $/mu$ and Standard Deviation $/sigma$.

The distribution function

and also its inverse (quantile) are difficult to compute. Let us consider the standardized variable

with the density $g$ and distribution function $G$, then

Figure 2.8: Density of the Standard Normal Distribution.

This distribution is also called standard normal distribution $N(0,1)$, and $G$ resp. the inverse is tabled in many text books. For the transformation we simply need

The model of the normal distribution is used in practice, if e.g. the investigated mineralization, as in iron-ore, has high grades and the variability is relatively small, as also in the thickness of stratified ore deposits or the density of ore.

Example 2.4: Let us consider the data in the example at page , and the model of normal distribution. How large are presumably the shortest intervals (tolerance intervals), which consist of approximately 70%, 90%, 95% or 99% of the values? From the table of probability distributions (see Appendix) we find

$/begin{array}{lll} Q_{.85} & = & 1.04// Q_{.95} & = & 1.64// Q_{.975}& = & 1.96// Q_{.995}& = & 2.58. /end{array}$

In case of a standard normal distributed random variable $Y$ we would get intervals of the sort [ $-1.04, 1.04], [-1.64, 1.64$], etc. From the data (grouped in the Stem and Leaf Display) the mean is estimated as $/bar{z}=23.5$ (as approximation of $/mu$) and the standard deviation $s=1.86$. Therefore the transformed intervals of

are computed to be:

$/begin{array}{llcl} 70 /% : & [-1.04 /times s + /bar{z}, 1.04 /times s + /bar{z... ....58 /times 1.86+23.5, 2.58 /times 1.86+23.5] & = & [18.7, 28.3].// /end{array}$