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 $\sigma=4$
$\therefore \quad \frac{\Sigma x_{i}}{100}=50$
$\Rightarrow \quad \Sigma x_{i}=5000$
$\text { and } \sigma^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\right)^{2}$
$\Rightarrow \quad (4)^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{100}-(50)^{2}$
$\Rightarrow \quad 16=\frac{\Sigma f_{i} x_{i}^{2}}{100}-2500$
$\Rightarrow \frac{\Sigma f_{i} x_{i}^{2}}{100}=16+2500=2516$
$\Sigma f_{i} x_{i}^{2}=251600$
Find the mean and variance for the first $10$ multiples of $3$
Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to
The mean and standard deviation of the marks of $10$ students were found to be $50$ and $12$ respectively. Later, it was observed that two marks $20$ and $25$ were wrongly read as $45$ and $50$ respectively. Then the correct variance is $............$.
Calculate the mean, variance and standard deviation for the following distribution:
Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
$f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
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