JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 12)
The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :
2
3
1
4
Explanation
$$\overline x = {{2 + 4 + + 10 + 12 + 14 + x + y} \over 7} = 8$$
x + y = 14 ....(i)
$${(\sigma )^2} = {{\sum {{{({x_i})}^2}} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2}$$
$$ \Rightarrow $$ $$16 = {{4 + 16 + 100 + 144 + 196 + {x^2} + {y^2}} \over 2} - {8^2}$$
$$ \Rightarrow $$ $$16 + 64 = {{460 + {x^2} + {y^2}} \over 7}$$
$$ \Rightarrow $$ 560 = 460 + x2 + y2
$$ \Rightarrow $$ x2 + y2 = 100 ......(ii)
Clearly by (i) and (ii),
(x + y)2 - 2xy = 100
$$ \Rightarrow $$ (14)2 - 2xy = 100
$$ \Rightarrow $$ 2xy = 96
$$ \Rightarrow $$ xy = 48
Now, |x - y| = $$\sqrt {{{\left( {x + y} \right)}^2} - 4xy} $$
= $$\sqrt {196 - 192} $$
= 2
x + y = 14 ....(i)
$${(\sigma )^2} = {{\sum {{{({x_i})}^2}} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2}$$
$$ \Rightarrow $$ $$16 = {{4 + 16 + 100 + 144 + 196 + {x^2} + {y^2}} \over 2} - {8^2}$$
$$ \Rightarrow $$ $$16 + 64 = {{460 + {x^2} + {y^2}} \over 7}$$
$$ \Rightarrow $$ 560 = 460 + x2 + y2
$$ \Rightarrow $$ x2 + y2 = 100 ......(ii)
Clearly by (i) and (ii),
(x + y)2 - 2xy = 100
$$ \Rightarrow $$ (14)2 - 2xy = 100
$$ \Rightarrow $$ 2xy = 96
$$ \Rightarrow $$ xy = 48
Now, |x - y| = $$\sqrt {{{\left( {x + y} \right)}^2} - 4xy} $$
= $$\sqrt {196 - 192} $$
= 2
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