If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and the total number of item is $18,$ find the mean and standard deviation.
Given, $n=18, \Sigma(x-5)=3$ and $\Sigma(x-5)^{2}=43$
$\therefore \quad \operatorname{Mean}=A+\frac{\Sigma(x-5)}{18}=5+\frac{3}{18}=5+0.1666=5.1666=5.17$
and $\quad SD =\sqrt{\frac{\Sigma(x-5)^{2}}{n}-\left(\frac{\Sigma(x-5)}{n}\right)^{2}}=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}$
$=\sqrt{2.3889-(0.166)^{2}}=\sqrt{2.3889-0.0277}=1.53$
If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ then the standard deviation of the $9$ items ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :
The mean and the standard deviation $(s.d.)$ of five observations are $9$ and $0,$ respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10,$ then their $s.d.$ is?
Find the mean and variance for the first $10$ multiples of $3$
Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to
The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
If it is replaced by $12$