JEE Main 2024 — Mathematics Probability & Statistics

Given: $\displaystyle \sum _{k=1}^{10}{a}_{k}=50$k=1∑10​ak​=50

$$\Rightarrow {a}_{1}+{a}_{2}+\ldots +{a}_{10}=50...(i)$$⇒a1​+a2​+…+a10​=50...(i)

$$\Rightarrow {({a}_{1}+{a}_{2}+\ldots +{a}_{10})}^{2}=2500$$⇒(a1​+a2​+…+a10​)2=2500

$$\Rightarrow \sum _{i=1}^{10}{a}_{i}^{2}+2\underset{k<j}{\sum }{a}_{k}{a}_{j}=2500$$⇒i=1∑10​ai2​+2k<j∑​ak​aj​=2500

$$\Rightarrow \sum _{i=1}^{10}{a}_{i}^{2}+2\times 1100=2500$$⇒i=1∑10​ai2​+2×1100=2500

$$\Rightarrow \sum _{i=1}^{10}{a}_{i}^{2}=2500-2(1100)$$⇒i=1∑10​ai2​=2500−2(1100)

$$\Rightarrow \sum _{i=1}^{10}{a}_{i}^{2}=300$$⇒i=1∑10​ai2​=300

Standard deviation, $\displaystyle \sigma =\sqrt{\frac{\sum {a}_{i}^{2}}{n}-{(\frac{\sum {a}_{i}}{n})}^{2}}$σ=n∑ai2​​−(n∑ai​​)2​

$$\Rightarrow \sigma =\sqrt{\frac{\sum {a}_{i}^{2}}{10}-{(\frac{\sum {a}_{i}}{10})}^{2}}$$⇒σ=10∑ai2​​−(10∑ai​​)2​

$$\Rightarrow \sigma =\sqrt{\frac{300}{10}-{(\frac{50}{10})}^{2}}$$⇒σ=10300​−(1050​)2​

$$\Rightarrow \sigma =\sqrt{30-25}=\sqrt{5}$$⇒σ=30−25​=5​