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(A) To find the class mark $(x_i)$ for each interval,the following relation is used:$x_i = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{

Vedclass · Accepted Answer

$145.20$

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(A) To find the class mark $(x_i)$ for each interval,the following relation is used:$x_i = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$Class size $(h)$ of this data $= 20$.Taking $150$ as the assumed mean $(a)$,$d_i$,$u_i$,and $f_i u_i$ can be calculated as follows:$\begin{array}{|c|c|c|c|c|c|} \hline 	ext{Daily wages} & f_i & x_i & d_i = x_i - 150 & u_i = \frac{d_i}{20} & f_i u_i \ \hline 100-120 & 12 & 110 & -40 & -2 & -24 \ \hline 120-140 & 14 & 130 & -20 & -1 & -14 \ \hline 140-160 & 8 & 150 & 0 & 0 & 0 \ \hline 160-180 & 6 & 170 & 20 & 1 & 6 \ \hline 180-200 & 10 & 190 & 40 & 2 & 20 \ \hline 	ext{Total} & 50 & & & & -12 \ \hline \end{array}$From the table,we have:$\sum f_i = 50$$\sum f_i u_i = -12$Using the step-deviation method,the mean $\bar{x}$ is given by:$\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} ight) 	imes h$$\bar{x} = 150 + \left( \frac{-12}{50} ight) 	imes 20$$\bar{x} = 150 - \frac{240}{50} = 150 - 4.8$$\bar{x} = 145.2$Therefore,the mean daily wage of the workers is $Rs. 145.20$.

Percentage of female teachers	$15$-$25$	$25$-$35$	$35$-$45$	$45$-$55$	$55$-$65$	$65$-$75$	$75$-$85$
Number of States/$U$.$T$.	$6$	$11$	$7$	$4$	$4$	$2$	$1$

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