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NCERT Class XI Mathematics - Statistics - Solutions

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Question : 14 of 34
Marks: +1, -0
First n natural numbers
Solution:  
Here xix_i = 1, 2, 3, 4, ..., n
∴ Σxi\Sigma x_i = 1 + 2 + 3 + 4 + ... + n = n(n+1)2\frac{n(n+1)}{2}
∴ Mean (x−)\left(\overset{-}{x}\right) = n(n+1)2n\frac{n(n+1)}{2n} = n+12\frac{n+1}{2}
Σxi2\Sigma x_i^2 = (1)2+(2)2+(3)2+(4)2(1)^2+(2)^2+(3)^2+(4)^2 + ... + n2n^2 = n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}
∴ Variance (σ2)(\sigma^2) =
n×n(n+1)(2n+1)6−[n(n+1)2]2n2\frac{n\times\frac{n(n+1)(2n+1)}{6} - \left[\frac{n(n+1)}{2}\right]^2}{n^2}
= (n+1)(2n+1)6\frac{(n+1)(2n+1)}{6} - (n+1)24\frac{(n+1)^2}{4}
= n+12(2n+13−n+12)\frac{n+1}{2}\left(\frac{2n+1}{3} - \frac{n+1}{2}\right) = n+12\frac{n+1}{2} (4n+2−3n+36)\left(\frac{4n+2-3n+3}{6}\right) = (n+1)(n−1)12\frac{(n+1)(n-1)}{12} = n2−112\frac{n^2-1}{12}
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