Mean and standard deviation of 100 items are 50 and $4,$ respectively. Then find the sum of all the item and the sum of the squares of the items.
Here, $\bar{x}=50, n=100$ and $\sigma=4$
$\therefore \quad \frac{\Sigma x_{i}}{100}=50$
$\Rightarrow \quad \Sigma x_{i}=5000$
$\text { and } \sigma^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\right)^{2}$
$\Rightarrow \quad (4)^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{100}-(50)^{2}$
$\Rightarrow \quad 16=\frac{\Sigma f_{i} x_{i}^{2}}{100}-2500$
$\Rightarrow \frac{\Sigma f_{i} x_{i}^{2}}{100}=16+2500=2516$
$\Sigma f_{i} x_{i}^{2}=251600$
The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three observations are $1, 2$ and $6$, the other two observations are
Suppose values taken by a variable $x$ are such that $a \le {x_i} \le b$, where ${x_i}$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ...n.$ Then..
The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals
Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$
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 :