Let a_1, a_2, ldots, a_{10} be 10 observation | vedclass

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(B) Given $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k < j} a_k a_j = 1100$. We know that $(\sum_{i=1}^{10} a_i)^2 = \sum_{i=1}^{10} a_i^2 + 2 \sum_{k < j}

Vedclass · Accepted Answer

$\sqrt{5}$

Vedclass · Answer

(B) Given $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k We know that $(\sum_{i=1}^{10} a_i)^2 = \sum_{i=1}^{10} a_i^2 + 2 \sum_{k Substituting the given values: $(50)^2 = \sum_{i=1}^{10} a_i^2 + 2(1100)$. $2500 = \sum_{i=1}^{10} a_i^2 + 2200$. $\sum_{i=1}^{10} a_i^2 = 2500 - 2200 = 300$. The variance $\sigma^2$ is given by $\frac{1}{n} \sum a_i^2 - (\bar{a})^2$,where $\bar{a} = \frac{\sum a_i}{n} = \frac{50}{10} = 5$. $\sigma^2 = \frac{300}{10} - (5)^2 = 30 - 25 = 5$. Therefore,the standard deviation $\sigma = \sqrt{5}$.

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

Let $a_1, a_2, \ldots, a_{10}$ be $10$ observations such that $\sum_{k=1}^{10} a_k = 50$ and $\sum_{k < j} a_k a_j = 1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to:

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