A data consists of n observations $x_1, x_2, \ldots ., x n$. If $\sum_{i=1}^n\left(x_i+1\right)^2=9 n$ and $\sum_{i=1}^n\left(x_i-1\right)^2=5 n$, then the standard deviation of this data is
Select the correct option:
A
$\sqrt{7}$
B
5
C
$\sqrt{5}$
D
2
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \sigma^2=\frac{1}{n} \sum_{i=1}^n x_i^2-\left(\frac{1}{n} \sum_{i=1}^n x_i\right)^2 \\ & \sigma^2=\frac{1}{n} A-\frac{1}{n^2} B^2 \\ & \sum_{i=1}^n\left(x_i+1\right)^2=9 n \\ & \Rightarrow A+n+2 B=9 n \Rightarrow A+2 B=8 n \\ & \sum_{i=1}^n\left(x_i+1\right)^2=5 n \\ & \Rightarrow A+n+2 B=5 n \Rightarrow A-2 B=4 n \\ & \Rightarrow \text { rom (ii) and (iii) } \\ & \quad A=6 n, B=n \\ & \quad \sigma^2=\frac{1}{n} \times 6 n-\frac{1}{n^2} \times n^2=6-1=5 \\ & \Rightarrow \quad \sigma=\sqrt{5}\end{aligned}$
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