JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 21)
Consider the statistics of two sets of observations as follows :
If the variance of the combined set of these two observations is $${{17} \over 9}$$, then the value of n is equal to ___________.
| Size | Mean | Variance | |
|---|---|---|---|
| Observation I | 10 | 2 | 2 |
| Observation II | n | 3 | 1 |
If the variance of the combined set of these two observations is $${{17} \over 9}$$, then the value of n is equal to ___________.
Answer
5
Explanation
For group - 1 : $${{\sum {{x_i}} } \over {10}} = 2 \Rightarrow \sum {{x_i}} = 20$$
$${{\sum {{x_i^2}} } \over {10}} - {(2)^2} = 2 \Rightarrow \sum {x_i^2} = 60$$
For group - 2 : $${{\sum {{y_i}} } \over n} = 3 \Rightarrow \sum {{y_i}} = 3n$$
$${{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \sum {y_i^2} = 10n$$
Now, combined variance
$${\sigma ^2} = {{\sum {\left( {x_i^2 + y_i^2} \right)} } \over {10 + n}} - {\left( {{{\sum {\left( {{x_i} + {y_i}} \right)} } \over {10 + n}}} \right)^2}$$
$$ \Rightarrow {{17} \over 9} = {{60 + 10n} \over {10 + n}} - {{{{(20 + 3n)}^2}} \over {{{(10 + n)}^2}}}$$
$$ \Rightarrow $$ 17 (n2 + 20n + 100) = 9(n2 + 40n + 200)
$$ \Rightarrow $$ 8n2 $$-$$ 20n $$-$$ 100 = 0
$$ \Rightarrow $$ 2n2 $$-$$ 5n $$-$$ 25 = 0 $$ \Rightarrow $$ n = 5
$${{\sum {{x_i^2}} } \over {10}} - {(2)^2} = 2 \Rightarrow \sum {x_i^2} = 60$$
For group - 2 : $${{\sum {{y_i}} } \over n} = 3 \Rightarrow \sum {{y_i}} = 3n$$
$${{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \sum {y_i^2} = 10n$$
Now, combined variance
$${\sigma ^2} = {{\sum {\left( {x_i^2 + y_i^2} \right)} } \over {10 + n}} - {\left( {{{\sum {\left( {{x_i} + {y_i}} \right)} } \over {10 + n}}} \right)^2}$$
$$ \Rightarrow {{17} \over 9} = {{60 + 10n} \over {10 + n}} - {{{{(20 + 3n)}^2}} \over {{{(10 + n)}^2}}}$$
$$ \Rightarrow $$ 17 (n2 + 20n + 100) = 9(n2 + 40n + 200)
$$ \Rightarrow $$ 8n2 $$-$$ 20n $$-$$ 100 = 0
$$ \Rightarrow $$ 2n2 $$-$$ 5n $$-$$ 25 = 0 $$ \Rightarrow $$ n = 5
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