JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 9)
Let in a series of 2n observations, half of them are equal to a and remaining half are equal to $$-$$a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a2 + b2 is equal to :
425
250
925
650
Explanation
Given series
(a, a, a, ........ n times), ($$-$$a, $$-$$a, $$-$$a, ...... n times)
Now $$\overline x $$ = $${{\sum {{x_i}} } \over {2n}} = 0$$
as, xi $$ \to $$ xi + b
then $$\overline x $$ $$ \to $$ $$\overline x $$ + b
So, $$\overline x $$ + b = 5 $$ \Rightarrow $$ b = 5
No change in S.D. due to change in origin
Standard deviation ($$\sigma$$) = $$\sqrt {{{\sum\limits_{i = 1}^{2n} {{{({x_i} - \overline x )}^2}} } \over {2n}}} $$
$$= \sqrt {{{\sum\limits_{i = 1}^{2n} {x_i^2} } \over {2n}}} = \sqrt {{{2n{a^2}} \over {2n}}} = \sqrt {{a^2}} $$
$$ \therefore $$ $$20 = \sqrt {{a^2}} \Rightarrow a = 20$$
$$ \therefore $$ a2 + b2 = 425
(a, a, a, ........ n times), ($$-$$a, $$-$$a, $$-$$a, ...... n times)
Now $$\overline x $$ = $${{\sum {{x_i}} } \over {2n}} = 0$$
as, xi $$ \to $$ xi + b
then $$\overline x $$ $$ \to $$ $$\overline x $$ + b
So, $$\overline x $$ + b = 5 $$ \Rightarrow $$ b = 5
No change in S.D. due to change in origin
Standard deviation ($$\sigma$$) = $$\sqrt {{{\sum\limits_{i = 1}^{2n} {{{({x_i} - \overline x )}^2}} } \over {2n}}} $$
$$= \sqrt {{{\sum\limits_{i = 1}^{2n} {x_i^2} } \over {2n}}} = \sqrt {{{2n{a^2}} \over {2n}}} = \sqrt {{a^2}} $$
$$ \therefore $$ $$20 = \sqrt {{a^2}} \Rightarrow a = 20$$
$$ \therefore $$ a2 + b2 = 425
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