Mean and standard deviation of 100 observations were found to be 40 and 10 , respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Given, $n=100, \bar{x}=40$ and $\sigma=10$
$\therefore \quad \frac{\Sigma x_{i}}{n}=40$
$\Rightarrow \quad \frac{\Sigma x_{i}}{100}=40$
$\Rightarrow \quad \Sigma x_{i}=4000$
Now, Corrected $\Sigma x_{i}=4000-30-70+3+27=3930$
Corrected mean $=\frac{2930}{100}=39.3$
Now, $\sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}=\frac{\Sigma x_{i}^{2}}{n}-(40)^{2}$
$\Rightarrow \quad 100=\frac{\Sigma x_{i}^{2}}{100}-1600$
$\Rightarrow \quad \Sigma x_{i}^{2}=170000$
Now, $\quad$ Corrected $\Sigma x_{i}^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}=164938$
Corrected $\sigma=\sqrt{\frac{164938}{100}-(39.3)^{2}}=\sqrt{1649.38-1544.49}=\sqrt{104.9}$
$=10.24$
The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ is
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 ?
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$, then the variance of the first four observations in equal to
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard deviation.
For a given distribution of marks mean is $35.16$ and its standard deviation is $19.76$. The co-efficient of variation is..