JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 11)
The mean and the standard deviation (s.d.) of
10 observations are 20 and 2 resepectively.
Each of these 10 observations is multiplied by
p and then reduced by q, where p $$ \ne $$ 0 and
q $$ \ne $$ 0. If the new mean and new s.d. become
half of their original values, then q is equal to
10
-20
-10
-5
Explanation
Let observations are
x1, x2, ...., x10
Here mean = 20 and standard deviation(S.D) = 2
When each of these 10 observations is multiplied by p then new observations are px1, px2, ....., px10
and new mean = 20p and new standard deviation(S.D) = 2|p|
Now when Reduced by q then new observations are
px1 - q, px2 - q, ....., px10 - q
and new mean = 20p - q and new standard deviation(S.D) = 2|p|
Given 20p - q = $${{20} \over 2}$$ = 10
and 2|p| = $${2 \over 2}$$ = 1
$$ \Rightarrow $$ p = $$ \pm $$ $${1 \over 2}$$
If p = $${1 \over 2}$$ then q = 0 (not possible as given q $$ \ne $$ 0)
If p = - $${1 \over 2}$$ then q = -20
Here mean = 20 and standard deviation(S.D) = 2
When each of these 10 observations is multiplied by p then new observations are px1, px2, ....., px10
and new mean = 20p and new standard deviation(S.D) = 2|p|
Now when Reduced by q then new observations are
px1 - q, px2 - q, ....., px10 - q
and new mean = 20p - q and new standard deviation(S.D) = 2|p|
Given 20p - q = $${{20} \over 2}$$ = 10
and 2|p| = $${2 \over 2}$$ = 1
$$ \Rightarrow $$ p = $$ \pm $$ $${1 \over 2}$$
If p = $${1 \over 2}$$ then q = 0 (not possible as given q $$ \ne $$ 0)
If p = - $${1 \over 2}$$ then q = -20
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