Let the mean of the data
$X$ | $1$ | $3$ | $5$ | $7$ | $9$ |
$(f)$ | $4$ | $24$ | $28$ | $\alpha$ | $8$ |
be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $..........$.
$7$
$6$
$8$
$5$
Let $ \bar x , M$ and $\sigma^2$ be respectively the mean, mode and variance of $n$ observations $x_1 , x_2,...,x_n$ and $d_i\, = - x_i - a, i\, = 1, 2, .... , n$, where $a$ is any number.
Statement $I$: Variance of $d_1, d_2,.....d_n$ is $\sigma^2$.
Statement $II$ : Mean and mode of $d_1 , d_2, .... d_n$ are $-\bar x -a$ and $- M - a$, respectively
The variance of the first $n$ natural numbers is
Let the mean and variance of the frequency distribution
$\mathrm{x}$ | $\mathrm{x}_{1}=2$ | $\mathrm{x}_{2}=6$ | $\mathrm{x}_{3}=8$ | $\mathrm{x}_{4}=9$ |
$\mathrm{f}$ | $4$ | $4$ | $\alpha$ | $\beta$ |
be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:
The $S.D$ of $15$ items is $6$ and if each item is decreased or increased by $1$, then standard deviation will be
The average marks of $10$ students in a class was $60$ with a standard deviation $4$ , while the average marks of other ten students was $40$ with a standard deviation $6$ . If all the $20$ students are taken together, their standard deviation will be