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
$5$
$3$
$4$
$0$
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 $S.D.$ of $5$ scores $1, 2, 3, 4, 5$ is
Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$
The mean and variance of $5$ observations are $5$ and $8$ respectively. If $3$ observations are $1,3,5$, then the sum of cubes of the remaining two observations is
For $(2n+1)$ observations ${x_1},\, - {x_1}$, ${x_2},\, - {x_2},\,.....{x_n},\, - {x_n}$ and $0$ where $x$’s are all distinct. Let $S.D.$ and $M.D.$ denote the standard deviation and median respectively. Then which of the following is always true