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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\sigma = \sqrt{5}$.

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:

Similar Questions

If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt{5}$ where $k > 0$,then $k$ is equal to

Let $x_1, x_2, \dots, x_n$ be the values taken by a variable $X$ and $y_1, y_2, \dots, y_n$ be the values taken by a variable $Y$ such that $y_i = ax_i + b$,where $i = 1, 2, \dots, n$. Then...

The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ are

Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$

If the variance of the data $2, 4, 5, 6, 8, 17$ is $23.33$,then the variance of $4, 8, 10, 12, 16, 34$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module