NTA Abhyas 2020
Exam NTA Abhyas
Year 2020
Subject Mathematics
Chapter Statistics NTA Abhyas 2020 · Mathematics · Statistics
Mean of the numbers 1,2, 3, , 9 with respective weights 1 ^ 2 + 1,...
Mean of the numbers
1 , 2 , 3 , … … … … , 9 1,2, 3, \ldots \ldots \ldots \ldots , 9 1 , 2 , 3 , ………… , 9 with respective weights
1 2 + 1 , 2 2 + 2 , 3 2 + 3 , … , 9 2 + 9 {1}^{2} + 1, {2}^{2} + 2, {3}^{2} + 3, \ldots , {9}^{2} + 9 1 2 + 1 , 2 2 + 2 , 3 2 + 3 , … , 9 2 + 9 is
Step-by-step solution
Here, for each
x i = i , {x}_{i} = i, x i = i , weight
= w i = i 2 + i = {w}_{i} = {i}^{2} + i = w i = i 2 + i Hence, the required mean
= Σ w i x i Σ w i = ∑ i = 1 9 i ( i 2 + i ) ∑ i = 1 9 ( i 2 + i ) = \dfrac{Σ{w}_{i}{x}_{i}}{Σ{w}_{i}} = \dfrac{\mathop{∑}_{i = 1}^{9}i\left({i}^{2} + i\right)}{\mathop{∑}_{i = 1}^{9}\left({i}^{2} + i\right)} = Σ w i Σ w i x i = ∑ i = 1 9 ( i 2 + i ) ∑ i = 1 9 i ( i 2 + i ) = ∑ i = 1 9 i 3 + ∑ i = 1 9 i 2 ∑ i = 1 9 i 2 + ∑ i = 1 9 i = 9 2 ( n + 1 ) 2 4 + 9 ( 9 + 1 ) ( 18 + 1 ) 6 9 ( 9 + 1 ) ( 18 + 1 ) 6 + 9 ( 9 + 1 ) 2 = 7 = \dfrac{\mathop{∑}_{i = 1}^{9}{i}^{3} + \mathop{∑}_{i = 1}^{9}{i}^{2}}{\mathop{∑}_{i = 1}^{9}{i}^{2} + \mathop{∑}_{i = 1}^{9}i} = \dfrac{\dfrac{{9}^{2}{\left(n + 1\right)}^{2}}{4} + \dfrac{9\left(9 + 1\right)\left(18 + 1\right)}{6}}{\dfrac{9\left(9 + 1\right)\left(18 + 1\right)}{6} + \dfrac{9\left(9 + 1\right)}{2}} = 7 = ∑ i = 1 9 i 2 + ∑ i = 1 9 i ∑ i = 1 9 i 3 + ∑ i = 1 9 i 2 = 6 9 ( 9 + 1 ) ( 18 + 1 ) + 2 9 ( 9 + 1 ) 4 9 2 ( n + 1 ) 2 + 6 9 ( 9 + 1 ) ( 18 + 1 ) = 7
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