Statistics Question 3
Question 3 - 2024 (27 Jan Shift 1)
Let $a_{1}, a_{2}, \ldots a_{10}$ be 10 observations such that $\sum_{\mathrm{k}=1}^{10} \mathrm{a}{\mathrm{k}}=50$ and $\sum{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}{\mathrm{k}} \cdot \mathrm{a}{\mathrm{j}}=1100$. Then the standard deviation of $a_{1}, a_{2}, \ldots, a_{10}$ is equal to :
(1) 5
(2) $\sqrt{5}$
(3) 10
(4) $\sqrt{115}$
Show Answer
Answer (2)
Solution
$\sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$
$a_{1}+a_{2}+\ldots+a_{10}=50 \ldots$
$\sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}{\mathrm{k}} \mathrm{a}{\mathrm{j}}=1100$.
If $\mathrm{a}{1}+\mathrm{a}{2}+\ldots+\mathrm{a}_{10}=50$.
$\left(a_{1}+a_{2}+\ldots+a_{10}\right)^{2}=2500$
$\Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}{\mathrm{i}}^{2}+2 \sum{\mathrm{k}<\mathrm{j}} \mathrm{a}{\mathrm{k}} \mathrm{a}{\mathrm{j}}=2500$
$\Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^{2}=2500-2(1100)$
$\sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^{2}=300$, Standard deviation ’ $\sigma$ '
$=\sqrt{\frac{\sum \mathrm{a}{\mathrm{i}}^{2}}{10}-\left(\frac{\sum \mathrm{a}{\mathrm{i}}}{10}\right)^{2}}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^{2}}$
$=\sqrt{30-25}=\sqrt{5}$
