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The mean and variance of eight observation are $9$ and $9.25$ respectively.If six of the observations are $6,7,10,12,12$ and $13$. Find the remaining two observations.

$\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)
$\therefore 9=\large\frac{\text{sum of all observation}}{8}$
=> $\large\frac{6+7+10+12+12+13+x+y}{8}$
=> $72 =60+x+y$
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
$x^2+y^2= 80$
Solve both the equations will give the value of x and y.
$x+y=12 => y=12-x$
Substituting the value of y in
$x^2+y^2= 80$
$2x^2-24 x +64=0$
divided by 2 we get,
$x(x-8) -4(x-8)=0$
$(x-4) (x-8) =0$
$x=4 \;or\;x=8$
When $x=4,x+y=12$
When $x=8,x+y=12$
The remaining two observations are 4 and 8
Hence A is the correct answer.
answered Jul 1, 2014 by meena.p

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