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(B) To find the mean,we first determine the class marks $(x_i)$ for each interval.Let the assumed mean $a = 50$ and class size $h = 10$. Then $d_i = x

Vedclass · Accepted Answer

$39.71$

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(B) To find the mean,we first determine the class marks $(x_i)$ for each interval.Let the assumed mean $a = 50$ and class size $h = 10$. Then $d_i = x_i - 50$ and $u_i = \frac{x_i - 50}{10}$.Percentage of female teachersNumber of states/$U$.$T$. $(f_i)$$x_i$$d_i = x_i - 50$$u_i = \frac{x_i - 50}{10}$$f_i x_i$$f_i d_i$$f_i u_i$$15$-$25$$6$$20$-$30$-$3$$120$-$180$-$18$$25$-$35$$11$$30$-$20$-$2$$330$-$220$-$22$$35$-$45$$7$$40$-$10$-$1$$280$-$70$-$7$$45$-$55$$4$$50$$0$$0$$200$$0$$0$$55$-$65$$4$$60$$10$$1$$240$$40$$4$$65$-$75$$2$$70$$20$$2$$140$$40$$4$$75$-$85$$1$$80$$30$$3$$80$$30$$3$Total$35$---$1390$-$360$-$36$From the table,we have $\Sigma f_i = 35, \Sigma f_i x_i = 1390, \Sigma f_i d_i = -360, \Sigma f_i u_i = -36$.$1$. Direct Method: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1390}{35} \approx 39.71$.$2$. Assumed Mean Method: $\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 50 + \frac{-360}{35} = 50 - 10.29 = 39.71$.$3$. Step-deviation Method: $\bar{x} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}ight) 	imes h = 50 + \left(\frac{-36}{35}ight) 	imes 10 = 50 - 10.29 = 39.71$.The mean percentage of female teachers is $39.71$.

Daily expenditure (in $Rs.$)	$100-150$	$150-200$	$200-250$	$250-300$	$300-350$
Number of households	$4$	$5$	$12$	$2$	$2$

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