The following table gives the literacy rate ( | vedclass

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(D) To find the class marks $(x_i)$,the following relation is used:$x_i = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$Class size $(h

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$69.43$

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(D) To find the class marks $(x_i)$,the following relation is used:$x_i = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$Class size $(h)$ for this data $= 10$.Taking $70$ as the assumed mean $(a)$,$d_i$,$u_i$,and $f_i u_i$ are calculated as follows:Literacy rate (in $\%$)Number of cities $(f_i)$$x_i$$d_i = x_i - 70$$u_i = \frac{d_i}{10}$$f_i u_i$$45-55$$3$$50$$-20$$-2$$-6$$55-65$$10$$60$$-10$$-1$$-10$$65-75$$11$$70$$0$$0$$0$$75-85$$8$$80$$10$$1$$8$$85-95$$3$$90$$20$$2$$6$Total$\sum f_i = 35$$\sum f_i u_i = -2$From the table,we obtain:$\sum f_i = 35$$\sum f_i u_i = -2$Mean,$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}ight) 	imes h$$= 70 + \left(\frac{-2}{35}ight) 	imes 10$$= 70 - \frac{20}{35}$$= 70 - \frac{4}{7}$$= 70 - 0.5714...$$\approx 69.43$Therefore,the mean literacy rate is $69.43 \%$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$2$	$4$	$10$	$20$	$18$	$20$	$16$	$10$

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.
Literacy rate (in $\%$) $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
Number of cities $3$ $10$ $11$ $8$ $3$

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Literacy rate (in $\%$)	$45-55$	$55-65$	$65-75$	$75-85$	$85-95$
Number of cities	$3$	$10$	$11$	$8$	$3$

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.Literacy rate (in $\%$)$45-55$$55-65$$65-75$$75-85$$85-95$Number of cities$3$$10$$11$$8$$3$