The frequency distribution of daily working expenditure of families in a locality is as follows:

Expenditure in $Rs. (x)$	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
No. of families $(f)$	$24$	$33$	$37$	$b$	$25$

If the mode of the distribution is $Rs. 140$,then the value of $b$ is:

The mean and standard deviation of a set of $n_{1}$ observations are $\bar{x}_{1}$ and $s_{1},$ respectively,while the mean and standard deviation of another set of $n_{2}$ observations are $\bar{x}_{2}$ and $s_{2},$ respectively. Show that the standard deviation of the combined set of $(n_{1}+n_{2})$ observations is given by $SD = \sqrt{\frac{n_{1}(s_{1})^{2}+n_{2}(s_{2})^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}(\bar{x}_{1}-\bar{x}_{2})^{2}}{(n_{1}+n_{2})^{2}}}$.

Let $a, b, c \in N$ and $a < b < c$. Let the mean,the mean deviation about the mean,and the variance of the $5$ observations $9, 25, a, b, c$ be $18, 4$,and $\frac{136}{5}$,respectively. Then $2a + b - c$ is equal to:

The mean and standard deviation of $100$ observations $x_1, x_2, \ldots, x_{100}$ were calculated as $40$ and $5.1$ respectively by a student who took by mistake $50$ instead of $40$ for one observation. Then the correct value of $\sum_{i=1}^{100} x_i^2=$

Let the mean and the variance of $20$ observations $x_{1}, x_{2}, \ldots, x_{20}$ be $15$ and $9$,respectively. For $\alpha \in R$,if the mean of $(x_{1}+\alpha)^{2}, (x_{2}+\alpha)^{2}, \ldots, (x_{20}+\alpha)^{2}$ is $178$,then the square of the maximum value of $\alpha$ is equal to $...........$

The mean and variance of eight observations are $9$ and $9.25,$ respectively. If six of the observations are $6, 7, 10, 12, 12,$ and $13,$ find the remaining two observations.