Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x<3 & 3 \leq x<5 & 5 \leq x<7 & 7 \leq x<10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$
$\begin{array}{|c|c|c|c|c|} \hline x & f _{ i } & x _{ i } & f x _{ i } & f x _{ i }^{ 2 } \\ \hline 1-3 & 6 & 2 & 12 & 24 \\ \hline 3-5 & 4 & 4 & 16 & 64 \\ \hline 5-7 & 5 & 6 & 30 & 180 \\ \hline 7-10 & 1 & 8.5 & 8.5 & 72.25 \\ \hline \text { Total } & n=16 & & \Sigma f_{i} x_{i}=66.5 & \Sigma f_{i}{ }_{i}^{2}=340.25 \\ \hline \end{array}$
$\therefore \quad$ Mean $=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{66.5}{16}=4.13$
And variance $=\sigma^{2}=\frac{\Sigma f_{i} x_{i}^{2}}{\Sigma f_{i}}-\left(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\right)^{2}=\frac{340.25}{16}-(4.13)^{2}$
$\quad=21.2656-17.0569=4.21$
The average marks of $10$ students in a class was $60$ with a standard deviation $4$ , while the average marks of other ten students was $40$ with a standard deviation $6$ . If all the $20$ students are taken together, their standard deviation will be
The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be
Find the mean and variance for the following frequency distribution.
Classes | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
Frequencies | $5$ | $8$ | $15$ | $16$ | $6$ |
Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is