While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Given $n=10, \bar{x}=45$ and $\sigma^{2}=16$
$\begin{array}{c}\bar{x}=45 \Rightarrow \frac{\Sigma x_{i}}{n}=45 \\\Rightarrow \quad \frac{\Sigma x_{i}}{10}=45 \Rightarrow \quad \Sigma x_{i}=450 \\\text { Corrected } \Sigma x_{i}=450-52+25=423\end{array}$
$\therefore \quad$ Corrected mean, $\bar{x}=\frac{423}{10}=42.3$
$\Rightarrow \quad \sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}$
$\begin{array}{ll}\Rightarrow & 16=\frac{\Sigma x_{i}^{2}}{10}-(45)^{2} \\ \Rightarrow & \Sigma x_{i}^{2}=20410\end{array}$
$\therefore \quad$ Corrected $\Sigma x_{i}^{2}=20410-(53)^{2}+(25)^{2}=18331$
And Corrected $\sigma^{2}=\frac{18331}{10}-(42.3)^{2}=43.81$
Consider three observations $a, b$ and $c$ such that $b = a + c .$ If the standard deviation of $a +2$ $b +2, c +2$ is $d ,$ then which of the following is true ?
Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to
The variance of the first $n$ natural numbers is
Consider a set of $3 n$ numbers having variance $4.$ In this set, the mean of first $2 n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3.$ A new set is constructed by adding $1$ into each of first $2 n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to .... .