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# The mean and variance of $7$ observation are $8$ and $16$ respectively.If six of the observations are $2,4,10,12,14$. Find the remaining two observations.

$\begin{array}{1 1}(A)\;4,8\\(B)\;5,8\\(C)\;6,8\\(D)\;5,7\end{array}$

Let us assume the remaining two observations to be x and y.
Given Mean $\bar {x} =8$
Variance= 16
Step 1:
We know that mean = $\large\frac{\text{sum of all observations}}{\text{number of observation}}$
Number of observations =7(Given)
$\therefore 8=\large\frac{\text{sum of all observation}}{7}$
$56=42+x+y$
$x+y=14$
Step 2:
We know the formula of variance
$\sigma^2= \large\frac{\sum x_i^2}{n} -\bigg( \large\frac{\sum x_i}{n}\bigg)^2$
Where by $\large\frac{\sum x_i}{n} $$=mean \therefore 16 =\large\frac{4+16+100+144+196+x^2+y^2}{7}$$-8^2$
$16=\large\frac{460+x^2+y^2}{7}$$-64$
$80=\large\frac{460+x^2+y^2}{7}$
$560= 460+x^2+y^2$
$x^2+y^2=100$
Step 3:
As calculated in step 1 and step 2
$x+y=14$
$x^2+y^2=100$
Solving both the equations will give the value of x and y.
$x+y=14 => y=14-x$
Substituting the value of y in
$x^2+y^2=100$
$x^2+(14-x)^2=100$
$2x^2+196+x^2-28x=100$
$2x^2-28x +96=0$
Divided by 2
$x^2-14x+48=0$
$x^2-8x-6x+48=0$
$x(x-8) -6(x-8)=0$
$(x-6) (x-8) =0$
$x=6;x=8$
When $x=6 => x+y=14$
$6+y=14$
$y=18$
When $x=8 => x+y=14$
$8+y=14$
$y=6$
The remaining two observations are 6 and 8
Hence C is the correct answer.