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.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Here, $\bar{x}=50, n=100$ and $\sigma=4$

$\therefore \quad \frac{\Sigma x_{i}}{100}=50$

$\Rightarrow \quad \Sigma x_{i}=5000$

$\text { and } \sigma^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\right)^{2}$

$\Rightarrow \quad (4)^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{100}-(50)^{2}$

$\Rightarrow \quad 16=\frac{\Sigma f_{i} x_{i}^{2}}{100}-2500$

$\Rightarrow \frac{\Sigma f_{i} x_{i}^{2}}{100}=16+2500=2516$

$\Sigma f_{i} x_{i}^{2}=251600$

Similar Questions

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

  • [JEE MAIN 2019]

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 $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

  • [AIEEE 2012]

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 :

  • [JEE MAIN 2024]