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BITSAT 2022 Paper

Section: Mathematics

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Question : 120 of 130

Marks: +1, -0

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1,2 and 6 , then the other two are

2 and 9
3 and 8
4 and 7
5 and 6

Solution: 👈: Video Solution

Let the two unknown items be

x

and

y

. Then,

Mean

= 4 \Rightarrow \frac{1+2+6+x+y}{5} = 4

\Rightarrow x+y=11 \text{ ...(i) }

and variance

=5.2

\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5} - (\text{mean})^2 = 5.2

\Rightarrow 41+x^2+y^2=5(5.2+16)

\Rightarrow 41+x^2+y^2 = 106

\Rightarrow x^2+y^2=65 \text{ ...(ii) }

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

x=4, y=7 \text{ or } x=7, y=4

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