Let the mean of the data

X
| vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha}$

$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Ri

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha}$

$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16$

$	ext { M.. }(\bar{x})=\frac{\sum f _{ i }\left| x _{ i }-\overline{ x }ight|}{\sum f _{ i }} 	ext { where } \sum f _{ i }=64+16=80$

$	ext { M.D. }(\bar{x})=\frac{4 	imes 4+24 	imes 2+28 	imes 0+16 	imes 2+8 	imes 4}{80}$

$=\frac{8}{5}$

$	ext { Variance }=\frac{\sum f _{ i }\left( x _{ i }-\overline{ x }ight)^2}{\sum f _{ i }}$

$=\frac{4 	imes 16+24 	imes 4+0+16 	imes 4+8 	imes 16}{80}=\frac{352}{80}$

$	herefore \frac{3 \alpha}{m+\sigma^2}=\frac{3 	imes 16}{\frac{128}{80}+\frac{352}{80}}=8$

$X$	$1$	$3$	$5$	$7$	$9$
$(f)$	$4$	$24$	$28$	$\alpha$	$8$

$\mathrm{x}$	$-2$	$-1$	$3$	$4$	$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$	$\frac{1}{5}$	$\mathrm{a}$	$\frac{1}{3}$	$\frac{1}{5}$	$\mathrm{~b}$

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

Similar Questions

Find the variance of the following data: $6,8,10,12,14,16,18,20,22,24$

Mean and variance of a set of $6$ terms is $11$ and $24$ respectively and the mean and variance of another set of $3$ terms is $14$ and $36$ respectively. Then variance of all $9$ terms is equal to

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

If $\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} - 8)^2 = 45} $ then the standard deviation of $x_1, x_2, ...... x_{18}$ is :-

Let $\mathrm{X}$ be a random variable with distribution.

$\mathrm{x}$ $-2$ $-1$ $3$ $4$ $6$

$\mathrm{P}(\mathrm{X}=\mathrm{x})$ $\frac{1}{5}$ $\mathrm{a}$ $\frac{1}{3}$ $\frac{1}{5}$ $\mathrm{~b}$

If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :

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