JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 2)
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :
10, 11
3, 18
8, 13
1, 20
Explanation
Let other two numbers be a, (21 $$-$$ a)
Now,
$$10.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}$$
(Using formula for variance)
$$ \Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}$$
$$ \Rightarrow {a^2} + {(21 - a)^2} = 221$$
$$\therefore$$ a = 10 and (21 $$-$$ a) = 21 $$-$$ 10 = 11
So, remaining two observations are 10, 11.
$$\Rightarrow$$ Option (1) is correct.
Now,
$$10.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}$$
(Using formula for variance)
$$ \Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}$$
$$ \Rightarrow {a^2} + {(21 - a)^2} = 221$$
$$\therefore$$ a = 10 and (21 $$-$$ a) = 21 $$-$$ 10 = 11
So, remaining two observations are 10, 11.
$$\Rightarrow$$ Option (1) is correct.
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