JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 19)
Let the observations xi (1 $$ \le $$ i $$ \le $$ 10) satisfy the
equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40.
If $$\mu $$ and $$\lambda $$ are the mean and the variance of the
observations, x1 – 3, x2 – 3, ...., x10 – 3, then
the ordered pair ($$\mu $$, $$\lambda $$) is equal to :
equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40.
If $$\mu $$ and $$\lambda $$ are the mean and the variance of the
observations, x1 – 3, x2 – 3, ...., x10 – 3, then
the ordered pair ($$\mu $$, $$\lambda $$) is equal to :
(6, 6)
(3, 3)
(3, 6)
(6, 3)
Explanation
$$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10
$$ \Rightarrow $$ x1 + x2 + .... + x10 = 60 ....(1)
$$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40
$$ \Rightarrow $$ ($$x_1^2 + x_2^2 + ... + x_{10}^2$$) + 25 $$ \times $$ 10 -
10( x1 + x2 + .... + x10) = 40
$$ \Rightarrow $$ $$x_1^2 + x_2^2 + ... + x_{10}^2$$ = 390 .....(2)
From question,
$$\mu $$ = $${{\left( {{x_1} - 3} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}$$
= $${{60 - 3 \times 10} \over {10}}$$ = 3
And $$\lambda $$ = variance = $${{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}$$ - $$\mu $$2
= $${{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}$$ - 9
= $${{390 + 90 - 360} \over {10}}$$ - 9
= 12 - 9 = 3
$$ \Rightarrow $$ x1 + x2 + .... + x10 = 60 ....(1)
$$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40
$$ \Rightarrow $$ ($$x_1^2 + x_2^2 + ... + x_{10}^2$$) + 25 $$ \times $$ 10 -
10( x1 + x2 + .... + x10) = 40
$$ \Rightarrow $$ $$x_1^2 + x_2^2 + ... + x_{10}^2$$ = 390 .....(2)
From question,
$$\mu $$ = $${{\left( {{x_1} - 3} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}$$
= $${{60 - 3 \times 10} \over {10}}$$ = 3
And $$\lambda $$ = variance = $${{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}$$ - $$\mu $$2
= $${{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}$$ - 9
= $${{390 + 90 - 360} \over {10}}$$ - 9
= 12 - 9 = 3
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