JEE Main 2026 — Mathematics Probability & Statistics

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

$$\Rightarrow \sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i + 40 = 180$$⇒i=1∑10​xi2​+4i=1∑10​xi​+40=180

$$\Rightarrow \sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i = 140$$⇒i=1∑10​xi2​+4i=1∑10​xi​=140

Also given $\displaystyle \sum_{i=1}^{10}(x_i-1)^2=90$i=1∑10​(xi​−1)2=90

$$\Rightarrow \sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i + 10 = 90$$⇒i=1∑10​xi2​−2i=1∑10​xi​+10=90

$$\Rightarrow \sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i = 80$$⇒i=1∑10​xi2​−2i=1∑10​xi​=80

Subtracting the second equation from the first:

$$6\sum_{i=1}^{10} x_i = 60 \Rightarrow \sum_{i=1}^{10} x_i = 10$$6i=1∑10​xi​=60⇒i=1∑10​xi​=10

Substituting this value in the second equation:

$$\sum_{i=1}^{10} x_i^2 - 2(10) = 80 \Rightarrow \sum_{i=1}^{10} x_i^2 = 100$$i=1∑10​xi2​−2(10)=80⇒i=1∑10​xi2​=100

The standard deviation is given by:

$$\sigma = \sqrt{\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2}$$σ=n∑xi2​​−(n∑xi​​)2​

$$\sigma = \sqrt{\dfrac{100}{10} - \left(\dfrac{10}{10}\right)^2}$$σ=10100​−(1010​)2​

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

Answer: $3$3