From a lot of $12$ items containing $3$ defectives, a sample of $5$ items is drawn at random. Let the random variable $\mathrm{X}$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$, where $\operatorname{gcd}(m, n)=1$, then $n-m$ is equal to..........

If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ then the standard deviation of the $9$ items  ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :

If the mean deviation about median for the number $3,5,7,2\,k , 12,16,21,24$ arranged in the ascending order, is $6$ then the median is

Given that $\bar{x}$ is the mean and $\sigma^{2}$ is the variance of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ Prove that the mean and variance of the observations $a x_{1}, a x_{2}, a x_{3}, \ldots ., a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2},$ respectively, $(a \neq 0)$

The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......

The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3,$ respectively. Later on it was found that three observations were incorrect, which were recorded as $21,21$ and $18 .$ Find the mean and standard deviation if the incorrect observations are omitted.