JEE 2024 — Mathematics PYQ

Since, ${\sum }_{k=1}^{10}{a}_k=50$  

$\Rightarrow \text{Mean }(X)=\frac{{\sum }_{k=1}^{10}{a}_k}{10}=\frac{50}{10}=5$  

Also,  

${\left({\sum }_{i=1}^{10}{a}_i\right)}^2={\sum }_{i=1}^{10}{a}_i^2+2\sum (a_i{a}_j)$

$\Rightarrow 2500={\sum }_{i=1}^{10}{a}_i^2+2\times 1100$  

$\Rightarrow {\sum }_{i=1}^{10}{a}_i^2=300$  

$\therefore {\sigma }^2=\frac{300}{10}-5^2=25$  

So, $\sigma =\sqrt{25}=5$