Question

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

• A. $\frac{nM-a}{4}$
• B. $\frac{nM+a}{2}$
• C. $\frac{nM-a}{2}$
• D. $nM+a$

Answer 1

The correct option is A $\frac{nM-a}{4}$

Let $n$ observation be, $x_1,x_2,x_3,...................x_n$
Given $\frac{x_1+x_2+x_3+...........+x_n}{n}=M..\left(1\right)$
Also given $x_1+x_2+x_3+....+x_{n-4}=a..\left(2\right)$
$\therefore x_{n-3}+x_{n-2}+x_{n-1}+x_n=nM-a$ (Using (1) and (2))
Hence mean of last four observation $=\frac{x_{n-3}+x_{n-2}+x_{n-1}+x_n}{4}=\frac{nM-a}{4}$