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KEAM 2015 Math Paper

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Question : 104 of 120

Marks: +1, -0

The mean of five observations is 4 and their variance is 5.2 .If three of these observations are 2,4 and 6, then the other two observations are

3 and 5
2 and 6
4 and 4
8 and 10
1 and 7

Solution:

Let the observations be

2,4,6, x, y

Now,

\overline{x}\;=\frac{\Sigma x_i}{N}

\Rightarrow 4=\;\frac{2+4+6+x+y}{5}

\Rightarrow \;\; x+y=8

.....(i)

Also,

\sigma^{2}=\;\frac{\Sigma x_i^{2}}{N}-(\overline{x})^{2}

\Rightarrow (5.2)^{2}=\;\frac{4+16+36+x^{2}+y^{2}}{5}-(4)^{2}

\Rightarrow \;\; x^{2}+y^{2}=50

.....(ii)

From Eqs. (i) and (ii), we get

x^{2}+(8-x)^{2}=50

\Rightarrow 2x^{2}-16x+64=50

\Rightarrow 2x^{2}-16x+14=0

\Rightarrow x^{2}-8x+7=0

\Rightarrow (x-1)(x-7)=0

\Rightarrow x=1,7

\Rightarrow y=7,1

Hence, remaining two observations are 1 and 7 .

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