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KEAM 2012 Math Paper

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Question : 104 of 120

Marks: +1, -0

The mean of five numbers is 0 and their variance is 2. If three of those numbers are -1, 1 and 2, then the other two numbers are

- 5 and 3
-4 and 2
-3 and 1
-2 and 0
-1 and -1

Solution:

Let the other two numbers be

x

and

y

.

\therefore

Mean

= \frac{-1+1+2+x+y}{5}

\Rightarrow\; 0 = \frac{2+x+y}{5}

\Rightarrow\; x+y=-2

.....(i)

and variance

= \frac{\Sigma (x_i - \overline{x})^2}{n}

\Rightarrow\; 2 = \frac{\left[(-1-0)^2 + (1-0)^2 + (2-0)^2 + (x-0)^2 + (y-0)^2\right]}{5}

\Rightarrow\; 10 = 1+1+4+x^2+y^2

\Rightarrow\; x^2+y^2=4

......(ii)

On squaring Eq. (i), we get

x^2+y^2+2xy=4

\Rightarrow\; 4+2xy=4

[From Eq. (ii)]

\Rightarrow\; xy=0

Now,

x-y = \sqrt{(x+y)^2 - 4xy}

= \sqrt{(-2)^2 - 4 \times 0} = 2

[From Eq. (i)]

\Rightarrow\; x-y=2

.....(iii)

On solving Eqs. (i) and (iii), we get

2x=0 \Rightarrow x=0

and

y=-2

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