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(A) Let the midpoints of the classes be $x_i = 15, 25, 35$. To simplify,shift the origin by $d_i = x_i - 25$,so $d_i = -10, 0, 10$. The frequencies ar

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$4$

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(A) Let the midpoints of the classes be $x_i = 15, 25, 35$. To simplify,shift the origin by $d_i = x_i - 25$,so $d_i = -10, 0, 10$. The frequencies are $f_i = 2, x, 2$. The mean $\bar{d} = \frac{\sum f_i d_i}{\sum f_i} = \frac{2(-10) + x(0) + 2(10)}{2+x+2} = \frac{0}{x+4} = 0$. The variance $\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} - (\bar{d})^2$. Given $\sigma^2 = 50$,we have $50 = \frac{2(-10)^2 + x(0)^2 + 2(10)^2}{x+4} - 0^2$. $50 = \frac{200 + 0 + 200}{x+4}$. $50 = \frac{400}{x+4}$. $x+4 = \frac{400}{50} = 8$. $x = 8 - 4 = 4$.

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

Subject	Mathematics,Physics,Chemistry
Mean	$42, 32, 40.9$
Standard deviation	$12, 15, 20$

\text{Particulars}	\text{Firm } $A$ \text{ and Firm } $B$
\text{No. of wage earners}	$A: 586, B: 648$
\text{Mean of monthly wages}	$A: Rs. 5253, B: Rs. 5253$
\text{Variance of wages}	$A: 100, B: 121$

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

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

Frequency $2, x, 2$

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