If the variance of the following frequency di | vedclass

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

Question

Variance is independent of shifting of origin

$\Rightarrow y_{i}: 15 \quad 25 \quad 35 \;\; or\;\;-10 \quad 0 \quad 10$

$\Rightarrow f_{i}: 2 \q

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Variance is independent of shifting of origin

$\Rightarrow y_{i}: 15 \quad 25 \quad 35 \;\; or\;\;-10 \quad 0 \quad 10$

$\Rightarrow f_{i}: 2 \quad \;\;\;x \quad \;2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2 \quad \;\;x \quad \;2$

$\Rightarrow \quad$ Variance $\left(\sigma^{2}\right)=\frac{\Sigma x _{ i }^{2} f _{ i }}{\Sigma f _{ i }}-(\overrightarrow{ x })^{2}$

$\Rightarrow \quad 50=\frac{200+0+200}{x+4}-0 \quad\{\bar{x}=0\}$

$\Rightarrow \quad 200+50 x=200+200$

$\Rightarrow \quad x=4$

Class	$10-20$	$20-30$	$30-40$
Frequency	$2$	$x$	$2$

	Height	Weight
Mean	$162.6\,cm$	$52.36\,kg$
Variance	$127.69\,c{m^2}$	$23.1361\,k{g^2}$

If the variance of the following frequency distribution is $50$ then $x$ is equal to:

Class $10-20$ $20-30$ $30-40$

Frequency $2$ $x$ $2$

Similar Questions

The variance of first $50$ even natural numbers is

The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$

Height Weight

Mean $162.6\,cm$ $52.36\,kg$

Variance $127.69\,c{m^2}$ $23.1361\,k{g^2}$

Can we say that the weights show greater variation than the heights?

If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is

The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...

The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is

If the variance of the following frequency distribution is $50$ then $x$ is equal to: Class $10-20$ $20-30$ $30-40$ Frequency $2$ $x$ $2$