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 variance of the first $n$ natural numbers is
One set containing five numbers has mean $8$ and variance $18$ and the second set containing $3$ numbers has mean $8$ and variance $24$. Then the variance of the combined set of numbers is
Find the mean and variance for the first $n$ natural numbers
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
If the variance of the first $n$ natural numbers is $10$ and the variance of the first m even natural numbers is $16$, then $m + n$ is equal to