Question

The A.M. of 'n' observations is M. If the sum of $(n-4)$ observation is 'a', what is the mean of remaining $4$ observations?

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

Answer 1

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

Given $\frac{a_1+a_2+a_3+a_4+....+a_n}{n}=M\left(1\right)(a_1+a_2+a_3+a_4+....+a_{n-4})=a\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$a+a_{n-3}+a_{n-2}+a_{n-1}+a_n=nM{a}_{n-3}+a_{n-2}+a_{n-1}+a_n=(nM-a)Mean=\frac{a_{n-3}+a_{n-2}+a_{n-1}+a_n}{4}=\frac{nM-a}{4}Mean=\frac{(nM-a)}{4}$