Statistics Question 3
Question 3 - 2024 (27 Jan Shift 1)
Let $a _1, a _2, \ldots a _{10}$ be 10 observations such that $\sum _{k=1}^{10} a _k=50$ and $\sum _{\forall k<j} a _k \cdot a _j=1100$. 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
$\sum _{k=1}^{10} a _k=50$
$a _1+a _2+\ldots+a _{10}=50 \ldots$
$\sum _{\forall k<j} a _k a _j=1100$.
If $a _1+a _2+\ldots+a _{10}=50$.
$\left(a _1+a _2+\ldots+a _{10}\right)^{2}=2500$
$\Rightarrow \sum _{i=1}^{10} a _i^{2}+2 \sum _{k<j} a _k a _j=2500$
$\Rightarrow \sum _{i=1}^{10} a _i^{2}=2500-2(1100)$
$\sum _{i=1}^{10} a _i^{2}=300$, Standard deviation ’ $\sigma$ '
$=\sqrt{\frac{\sum a _i^{2}}{10}-\left(\frac{\sum a _i}{10}\right)^{2}}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^{2}}$
$=\sqrt{30-25}=\sqrt{5}$
