Suppose a class has $7$ students. The average marks of these students in the mathematics examination is $62$, and their variance is $20$ . A student fails in the examination if $he/she$ gets less than $50$ marks, then in worst case, the number of students can fail is

Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :

The varience of data $1001, 1003, 1006, 1007, 1009, 1010$ is -

Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are co-prime, then $m + n$ is equal to

Let the mean and standard deviation of marks of class $A$ of $100$ students be respectively $40$ and $\alpha( > 0)$, and the mean and standard deviation of marks of class $B$ of $n$ students be respectively $55$ and $30-\alpha$. If the mean and variance of the marks of the combined class of $100+ n$ students are respectively $50$ and $350$,then the sum of variances of classes $A$ and $B$ is