Q33 / 75 · JEE Main 2026 (04 April Shift 2) · Mathematics

For $10$ observations $x_1, x_2, \ldots, x_{10}$, if $\sum_{i=1}^{10}(x_i+2)^2=180$ and $\sum_{i=1}^{10}(x_i-1)^2=90$, then their standard deviation is:

  • A
    $2$
  • B
    $\sqrt{3}$
  • C
    $2\sqrt{2}$
  • Correct answer
    D
    $3$

Solution

Given $\sum_{i=1}^{10}(x_i+2)^2=180$

$\Rightarrow \sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i + 40 = 180$

$\Rightarrow \sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i = 140$

Also given $\sum_{i=1}^{10}(x_i-1)^2=90$

$\Rightarrow \sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i + 10 = 90$

$\Rightarrow \sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i = 80$

Subtracting the second equation from the first:

$6\sum_{i=1}^{10} x_i = 60 \Rightarrow \sum_{i=1}^{10} x_i = 10$

Substituting this value in the second equation:

$\sum_{i=1}^{10} x_i^2 - 2(10) = 80 \Rightarrow \sum_{i=1}^{10} x_i^2 = 100$

The standard deviation is given by:

$\sigma = \sqrt{\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2}$

$\sigma = \sqrt{\dfrac{100}{10} - \left(\dfrac{10}{10}\right)^2}$

$\sigma = \sqrt{10 - 1} = \sqrt{9} = 3$

Answer: $3$

Hence, the correct answer is option: 'D'.

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