The AM of n observations is M. If the sum of (n - 4) observations is a ,then the mean of remaining four observations is 

Solution:

Let ${\mathrm{x}}_1,{\mathrm{x}}_2,\dots ,{\mathrm{x}}_{\mathrm{n}}$ be n observations.

$\begin{array}{l}\begin{array}{rl}\therefore \mathrm{M}& =\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+\dots +{\mathrm{x}}_{\mathrm{n}}}{\mathrm{n}}\\ & =\frac{{\mathrm{x}}_1+{\mathrm{x}}_2+\dots +{\mathrm{x}}_{\mathrm{n}-4}+{\mathrm{x}}_{\mathrm{n}-3}+{\mathrm{x}}_{\mathrm{n}-2}+{\mathrm{x}}_{\mathrm{n}-1}+\mathrm{x}}{\mathrm{n}}\end{array}\\ \Rightarrow \mathrm{nM}=\mathrm{a}+{\mathrm{x}}_{\mathrm{n}-3}+{\mathrm{x}}_{\mathrm{n}-2}+{\mathrm{x}}_{\mathrm{n}-1}+{\mathrm{x}}_{\mathrm{n}}\\ \Rightarrow \frac{\mathrm{nM}-\mathrm{a}}{4}=\frac{{\mathrm{x}}_{\mathrm{n}-3}+{\mathrm{x}}_{\mathrm{n}-2}+{\mathrm{x}}_{\mathrm{n}-1}+{\mathrm{x}}_{\mathrm{n}}}{4}\end{array}$