JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 10)
If both the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 – 4)2
, (x2 – 4)2
,....., (x50 – 4)2
is :
400
480
380
525
Explanation
$$Mean(\mu ) = {{\sum {{x_i}} } \over {50}} = 16$$
$$ \therefore $$ $$\sum {{x_i}} = 16 \times 50$$
$$S.D.\left( \sigma \right) = \sqrt {{{\sum {{x_i}^2} } \over {50}} - {{\left( \mu \right)}^2}} = 16$$
$$ \Rightarrow {{\sum {{x_i}^2} } \over {50}} = 256 \times 2$$
Required mean = $${{\sum {{{\left( {{x_i} - 4} \right)}^2}} } \over {50}}$$
$$ \Rightarrow {{\sum {{x_i}^2} + 16 \times 50 - 8\sum {{x_i}} } \over {50}}$$
$$ \Rightarrow $$ 256 $$ \times $$ 2 + 16 - 8 $$ \times $$ 16
$$ \Rightarrow $$ 400
$$ \therefore $$ $$\sum {{x_i}} = 16 \times 50$$
$$S.D.\left( \sigma \right) = \sqrt {{{\sum {{x_i}^2} } \over {50}} - {{\left( \mu \right)}^2}} = 16$$
$$ \Rightarrow {{\sum {{x_i}^2} } \over {50}} = 256 \times 2$$
Required mean = $${{\sum {{{\left( {{x_i} - 4} \right)}^2}} } \over {50}}$$
$$ \Rightarrow {{\sum {{x_i}^2} + 16 \times 50 - 8\sum {{x_i}} } \over {50}}$$
$$ \Rightarrow $$ 256 $$ \times $$ 2 + 16 - 8 $$ \times $$ 16
$$ \Rightarrow $$ 400
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