$\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.

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