$\begin{array}{1 1}(A)\;4,8\\(B)\;5,8\\(C)\;3,8\\(D)\;5,7\end{array} $

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Let us assume the remaining two observations to be x and y.

Given Mean $\bar {x} =9$

Variance= 9.25

Step 1:

We know that mean = $\large\frac{\text{sum of all observations}}{\text{number of observation}}$

Number of observations =8(Given)

$\therefore 9=\large\frac{\text{sum of all observation}}{8}$

=> $\large\frac{6+7+10+12+12+13+x+y}{8}$

=> $72 =60+x+y$

$x+y=12$

Step 2:

we know the formula for variance(ie)

$\sigma^2= \large\frac{\sum x_i^2}{n} -\bigg( \large\frac{\sum x_i}{n}\bigg)^2$

Whereby $\large\frac{\sum x_i}{n} $$=mean$

$\therefore 9.25 =\large\frac{36+49+100+144+144+169+x^2+y^2}{8}$$-9^2$

=> $ 9.25 =\large\frac{642+ x^2+y^2}{8}$$-81$

=> $90.25 = \large\frac{642+x^2+y^2}{8}$

=> $ 722 = 642+x^2+y^2$

=> $ x^2+y^2=80$

Step 3:

As calculated in step 1 and step 2

$x+y=12$

$x^2+y^2= 80$

Solve both the equations will give the value of x and y.

$x+y=12 => y=12-x$

Substituting the value of y in

$x^2+y^2= 80$

$x^2+(12-x)^2=80$

$x^2+144+x^2-24x=80$

$2x^2-24x+144=80$

$2x^2-24 x +64=0$

divided by 2 we get,

$x^2-12x+32=0$

$x^2-8x-4x+32=0$

$x(x-8) -4(x-8)=0$

$(x-4) (x-8) =0$

$x=4 \;or\;x=8$

When $x=4,x+y=12$

$4+y=12$

$y=8$

When $x=8,x+y=12$

$8+y=12$

$y=4$

The remaining two observations are 4 and 8

Hence A is the correct answer.

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