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Statistics Part 2

Section: Mathematics

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Question : 6 of 50

Marks: +1, -0

For 10 observations

x_{1}, x_{2}, \dots, x_{10}

\sum\limits_{i=1}^{10} (x_{i}+2)^{2}=180

\sum\limits_{i=1}^{10} (x_{i}-1)^{2}=90

, then their standard deviation is :

[JEE Main 4 Apr 2026 Shift 2]

2
$\sqrt{3}$
$2 \sqrt{2}$
3

Solution:

\sum\limits_{i=1}^{10} (x_{i}+2)^{2}=180

\sum\limits_{i=1}^{10} x_i^2 + 4 \sum\limits_{i=1}^{10} x_i + \sum\limits_{i=1}^{10} 4 = 180

\sum\limits_{i=1}^{10} x_i^2 + 4 \sum\limits_{i=1}^{10} x_i = 180-40 \;\; ...(1)

Also

\sum\limits_{i=1}^{10} (x_{i}-1)^{2}=90

\sum\limits_{i=1}^{10} x_i^{2} - 2 \sum\limits_{i=1}^{10} x_i + \sum\limits_{i=1}^{10} 1 = 90

\sum\limits_{i=1}^{10} x_i^{2} - 2 \sum\limits_{i=1}^{10} x_i = 90-10 \;\; ...(2)

\sum\limits_{i=1}^{10} x_i^{2} + 4 \sum\limits_{i=1}^{10} x_i = 140

\sum\limits_{i=1}^{10} x_i^{2} - 2 \sum\limits_{i=1}^{10} x_i = 80

\sum\limits_{i=1}^{10} x_i^2 = 100 \text{ and } \sum\limits_{i=1}^{10} x_i = 10

\sigma^2 = \frac{ \sum\limits_{i=1}^{10} x_i^2 }{N} - \left( \frac{ \sum\limits_{i=1}^{10} x_i }{N} \right)^2

= \frac{100}{10} - \left( \frac{10}{10} \right)^2

\sigma^{2} = 10 - 1 = 9

\Rightarrow \sigma = 3

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