JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 11)
The mean and variance of seven observations are
8 and 16, respectively. If 5 of the observations are
2, 4, 10, 12, 14, then the product of the remaining
two observations is :
40
48
49
45
Explanation
Given mean ($$\mu $$) = 8
variance ($${\sigma ^2}$$) = 16
No of observations (N) = 7
Let the two unknown observation = x and y
We know,
$${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$$ = 16
$$ \Rightarrow $$ $${{{2^2} + {4^2} + {{10}^2} + {{12}^2} + {{14}^2} + {x^2} + {y^2}} \over 7} - {\left( 8 \right)^2}$$ = 16
$$ \Rightarrow $$ x2 + y2 = 100 ........(1)
We know,
$$\mu $$ = $${{\sum {{x_i}} } \over N}$$ = 8
$$ \Rightarrow $$ $${{2 + 4 + 10 + 12 + 14 + x + y} \over 7}$$ = 8
$$ \Rightarrow $$ x + y = 14 ...........(2)
As (x + y)2 = x2 + y2 + 2xy
$$ \Rightarrow $$ (14)2 = 100 + 2xy
$$ \Rightarrow $$ 196 = 100 + 2xy
$$ \Rightarrow $$ xy = 48
variance ($${\sigma ^2}$$) = 16
No of observations (N) = 7
Let the two unknown observation = x and y
We know,
$${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$$ = 16
$$ \Rightarrow $$ $${{{2^2} + {4^2} + {{10}^2} + {{12}^2} + {{14}^2} + {x^2} + {y^2}} \over 7} - {\left( 8 \right)^2}$$ = 16
$$ \Rightarrow $$ x2 + y2 = 100 ........(1)
We know,
$$\mu $$ = $${{\sum {{x_i}} } \over N}$$ = 8
$$ \Rightarrow $$ $${{2 + 4 + 10 + 12 + 14 + x + y} \over 7}$$ = 8
$$ \Rightarrow $$ x + y = 14 ...........(2)
As (x + y)2 = x2 + y2 + 2xy
$$ \Rightarrow $$ (14)2 = 100 + 2xy
$$ \Rightarrow $$ 196 = 100 + 2xy
$$ \Rightarrow $$ xy = 48
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