Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in R$ $\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is. . . . .

Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. 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 $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to ..............

The variance of $10$ observations is $16$. If each observation is doubled, then standard deviation of new data will be -

The mean of two samples of size $200$ and $300$ were found to be $25, 10$ respectively their $S.D.$ is $3$ and $4$ respectively then variance of combined sample  of size $500$ is :-

For the frequency distribution :

where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and

$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be 

Let $n \geq 3$. A list of numbers $x_1, x, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers $y_1, y_2, \ldots, y_n$ is made as follows $y_1=\frac{x_1+x_2}{2}, y_2=\frac{x_1+x_2}{2}$ and $y_j=x_j$ for $j=3,4, \ldots, n$.

The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Then, which of the following is necessarily true?