JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 20)
Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to __________.
Answer
68
Explanation
Let first 2n observations are x1, x2 ...................., x2n
and last n observations are y1, y2 ....................., yn
Now, $${{\sum {{x_i}} } \over {2n}} = 6$$, $${{\sum {{y_i}} } \over n} = 3$$
$$ \Rightarrow \sum {{x_i}} = 12n,\sum {{y_i}} = 3n$$
$$ \therefore $$ $${{\sum {{x_i}} + \sum {{y_i}} } \over {3n}} = {{15n} \over {3n}} = 5$$
Now, $${{\sum {x_i^2} + \sum {y_i^2} } \over {3n}} - {5^2} = 4$$
$$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n = 87n$$
Now, mean is $${{\sum {({x_i} + 1) + \sum {({y_i} - 1)} } } \over {3n}} = {{15n + 2n - n} \over {3n}} = {{16} \over 3}$$
Now, variance is $${{{{\sum {{{({x_i} + 1)}^2} + \sum {({y_i} - 1)} } }^2}} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$
$$ = {{\sum {x_i^2 + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n} } \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$
$$ = {{87n + 2(9n) + 3n} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$
= $$29 + 6 + 1 - {\left( {{{16} \over 3}} \right)^2}$$
$$ = {{324 - 256} \over 9} = {{68} \over 9} = k$$
$$ \Rightarrow $$ 9k = 68
Therefore, the correct answer is 68.
and last n observations are y1, y2 ....................., yn
Now, $${{\sum {{x_i}} } \over {2n}} = 6$$, $${{\sum {{y_i}} } \over n} = 3$$
$$ \Rightarrow \sum {{x_i}} = 12n,\sum {{y_i}} = 3n$$
$$ \therefore $$ $${{\sum {{x_i}} + \sum {{y_i}} } \over {3n}} = {{15n} \over {3n}} = 5$$
Now, $${{\sum {x_i^2} + \sum {y_i^2} } \over {3n}} - {5^2} = 4$$
$$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n = 87n$$
Now, mean is $${{\sum {({x_i} + 1) + \sum {({y_i} - 1)} } } \over {3n}} = {{15n + 2n - n} \over {3n}} = {{16} \over 3}$$
Now, variance is $${{{{\sum {{{({x_i} + 1)}^2} + \sum {({y_i} - 1)} } }^2}} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$
$$ = {{\sum {x_i^2 + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n} } \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$
$$ = {{87n + 2(9n) + 3n} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$
= $$29 + 6 + 1 - {\left( {{{16} \over 3}} \right)^2}$$
$$ = {{324 - 256} \over 9} = {{68} \over 9} = k$$
$$ \Rightarrow $$ 9k = 68
Therefore, the correct answer is 68.
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