The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.
$\begin{array}{|c|c|c|c|} \hline x & f_{i} & f_{1} x_{i} & f x_{i}^{2} \\ \hline A & 2 & 2 A & 2 A^{2} \\ \hline 2 A & 1 & 2 A & 4 A^{2} \\ \hline 3 A & 1 & 3 A & 9 A^{2} \\ \hline 4 A & 1 & 4 A & 16 A^{2} \\ \hline 5 A & 1 & 5 A & 25 A^{2} \\ \hline 6 A & 1 & 6 A & 36 A^{2} \\ \hline \text { Total } & n=7 & \Sigma f_{i}=22 A & \Sigma f_{i}^{2}=92 A^{2} \\ \hline \end{array}$
$\therefore \quad \sigma^{2}=\frac{\Sigma f_{t} x_{1}^{2}}{n}-\left(\frac{\Sigma f_{1} x_{1}}{n}\right)^{2}$
$\Rightarrow \quad 160=\frac{92 A^{2}}{7}-\left(\frac{22 A}{7}\right)^{2} \Rightarrow 160=\frac{92 A^{2}}{7}-\frac{484 A^{2}}{49}$
$\Rightarrow \quad 160=(644-484) \frac{A^{2}}{49} \Rightarrow 160=\frac{160 A^{2}}{49}$
$\Rightarrow \quad A^{2}=49 \quad \therefore \quad A=7$
If mean and standard deviation of $5$ observations $x_1 ,x_2 ,x_3 ,x_4 ,x_5$ are $10$ and $3$, respectively, then the variance of $6$ observations $x_1 ,x_2 ,.....,x_3$ and $-50$ is equal to
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
Height | Weight | |
Mean | $162.6\,cm$ | $52.36\,kg$ |
Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?
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 :