If the variance of the frequency distribution | vedclass

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(A) Let $A = 5$. We calculate $d_i = x_i - A$ and the required sums:						$x_i$			$f_i$			$d_i = x_i - 5$			$f_i d_i^2$			$f_i d_i$							$2$			$3$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let $A = 5$. We calculate $d_i = x_i - A$ and the required sums:						$x_i$			$f_i$			$d_i = x_i - 5$			$f_i d_i^2$			$f_i d_i$							$2$			$3$			$-3$			$27$			$-9$							$3$			$6$			$-2$			$24$			$-12$							$4$			$16$			$-1$			$16$			$-16$							$5$			$\alpha$			$0$			$0$			$0$							$6$			$9$			$1$			$9$			$9$							$7$			$5$			$2$			$20$			$10$							$8$			$6$			$3$			$54$			$18$			Total frequency $N = \sum f_i = 3 + 6 + 16 + \alpha + 9 + 5 + 6 = 45 + \alpha$.Sum $\sum f_i d_i = -9 - 12 - 16 + 0 + 9 + 10 + 18 = 0$.Sum $\sum f_i d_i^2 = 27 + 24 + 16 + 0 + 9 + 20 + 54 = 150$.Variance $\sigma^2 = \frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2 = 3$.$\frac{150}{45 + \alpha} - 0 = 3$.$150 = 3(45 + \alpha) \Rightarrow 150 = 135 + 3\alpha$.$3\alpha = 15 \Rightarrow \alpha = 5$.

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

If the variance of the frequency distribution is $3$,then $\alpha$ is ......

$X_i$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

Frequency $f_i$ $3$ $6$ $16$ $\alpha$ $9$ $5$ $6$

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$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

If the variance of the frequency distribution is $3$,then $\alpha$ is ...... $X_i$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ Frequency $f_i$ $3$ $6$ $16$ $\alpha$ $9$ $5$ $6$