91Ó°ÊÓ

$\begin{array}{r}{\mathrm{μ}}_{\mathrm{X}}=2.1\\ {\mathrm{σ}}_{\mathrm{X}}=1.136\\ {\mathrm{μ}}_{\mathrm{Y}}=1.0\\ {\mathrm{σ}}_{\mathrm{Y}}=1.0\end{array}$

X = quantity of nonword errors in an essay chosen at random

Y = quantity of word errors in an essay chosen at random

$\begin{array}{c}{\mathrm{μ}}_{\mathrm{NW}}=3×{\mathrm{μ}}_{\mathrm{X}}\\ =3×2.1\\ =6.3\\ {\mathrm{μ}}_{\mathrm{W}}=2×{\mathrm{μ}}_{\mathrm{Y}}\\ =2×1.0\\ =2.0\\ \mathrm{μ}={\mathrm{μ}}_{\mathrm{NW}}+{\mathrm{μ}}_{\mathrm{W}}\\ =6.3+2.0\\ =8.3\end{array}$

If every data point for non-word error is multiplied by 3, the distribution's centre is also multiplied by 3, so the measure of centre must be multiplied by 3, and the mean equals the measure of centre. Similarly, if every data point for word error is multiplied by 2, the distribution's centre is multiplied by 2, therefore the measure of centre is also multiplied by 2, and the mean is the measure of centre. The sum of the two random variables' means will be the mean of the two random variables. The total number of non-word and word errors is subtracted from the total average of 8.3 points.