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 $..........$.

  • [JEE MAIN 2023]
  • A

    $7$

  • B

    $6$

  • C

    $8$

  • D

    $5$

Similar Questions

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

  • [JEE MAIN 2014]

The variance of the first $n$ natural numbers is

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$\mathrm{x}$ $\mathrm{x}_{1}=2$ $\mathrm{x}_{2}=6$ $\mathrm{x}_{3}=8$ $\mathrm{x}_{4}=9$
$\mathrm{f}$ $4$ $4$ $\alpha$ $\beta$

be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be:

  • [JEE MAIN 2021]

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