If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to
$20$
$25$
$23$
$21$
The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five observations are $2, 4, 10,12,14,$ then the absolute difference of the remaining two observations is
The data is obtained in tabular form as follows.
${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
The $S.D$ of $15$ items is $6$ and if each item is decreased or increased by $1$, then standard deviation will be
Let sets $A$ and $B$ have $5$ elements each. Let the mean of the elements in sets $A$ and $B$ be $5$ and $8$ respectively and the variance of the elements in sets $A$ and $B$ be $12$ and $20$ respectively $A$ new set $C$ of $10$ elements is formed by subtracting $3$ from each element of $A$ and adding 2 to each element of B. Then the sum of the mean and variance of the elements of $C$ is $.......$.
The varience of data $1001, 1003, 1006, 1007, 1009, 1010$ is -