Let the observations mathrm{x}_{mathrm{i}}(1 | vedclass

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

Question

$\sum_{i=1}^{10}\left(x_{i}-5\right)=10$

$\Rightarrow$ Mean of observation $\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{3}\left(\ma

Vedclass · Accepted Answer

$(3, 3)$

Vedclass · Answer

$\sum_{i=1}^{10}\left(x_{i}-5ight)=10$

$\Rightarrow$ Mean of observation $\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{3}\left(\mathrm{x}_{\mathrm{i}}-5ight)=1$

$\Rightarrow \mu=$ mean of observation $\left(\mathrm{x}_{\mathrm{i}}-3ight)$

$=\left(	ext { mean of observation }\left(\mathrm{x}_{\mathrm{i}}-5ight)ight)+2$

$=1+2=3$

Variance of observation

$\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-5ight)^{2}-\left(\mathrm{Mean} 	ext { of }\left(\mathrm{x}_{\mathrm{i}}-5ight)ight)^{2}=3$

$\Rightarrow \quad \lambda=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-3ight)$

$=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-5ight)=3$ $	herefore \quad(\mu, \lambda)=(3,3)$

Class:	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

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

Similar Questions

If the mean of the frequency distribution

Class: $0-10$ $10-20$ $20-30$ $30-40$ $40-50$

Frequency $2$ $3$ $x$ $5$ $4$

is $28$ , then its variance is $........$.

If the standard deviation of the numbers $ 2,3,a $ and $11$ is $3.5$ then which of the following is true ?

If the mean and the variance of $6,4, a, 8, b, 12,10, 13$ are $9$ and $9.25$ respectively, then $a+b+a b$ 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 $..........$.

If the data $x_1, x_2, ...., x_{10}$ is such that the mean of first four of these is $11$, the mean of the remaining six is $16$ and the sum of squares of all of these is $2,000$; then the standard deviation of this data is