Verified Solution • 2025–26 Session moderate

QJEEM21Maf51c: JEE Main 2021 Mathematics - Class 11 JEEM21Maf51c Solution: Consider the statistics of two sets of observations as follows…

Consider the statistics of two sets of observations as follows :

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 179{{17} \over 9} , then the value of n is equal to ___________.

Step-by-Step Verified Solution

For group - 1 : ∑xi10=2⇒∑xi=20{{\sum {{x_i}} } \over {10}} = 2 \Rightarrow \sum {{x_i}} = 20

∑xi210−(2)2=2⇒∑xi2=60{{\sum {{x_i^2}} } \over {10}} - {(2)^2} = 2 \Rightarrow \sum {x_i^2} = 60

For group - 2 : ∑yin=3⇒∑yi=3n{{\sum {{y_i}} } \over n} = 3 \Rightarrow \sum {{y_i}} = 3n

∑yi2n−32=1⇒∑yi2=10n{{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \sum {y_i^2} = 10n

Now, combined variance

σ2=∑(xi2+yi2)10+n−(∑(xi+yi)10+n)2{\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}

⇒179=60+10n10+n−(20+3n)2(10+n)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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