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

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$8.1$

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(D) To find the class mark $(x_i)$ for each interval,the following relation is used:Class mark $(x_i) = \frac{	ext{Upper class limit} + 	ext{Lower class limit}}{2}$The values of $x_i$ and $f_ix_i$ are calculated as follows:Number of plantsNumber of houses $(f_i)$$x_i$$f_ix_i$$0-2$$1$$1$$1 	imes 1 = 1$$2-4$$2$$3$$2 	imes 3 = 6$$4-6$$1$$5$$1 	imes 5 = 5$$6-8$$5$$7$$5 	imes 7 = 35$$8-10$$6$$9$$6 	imes 9 = 54$$10-12$$2$$11$$2 	imes 11 = 22$$12-14$$3$$13$$3 	imes 13 = 39$Total$\sum f_i = 20$$\sum f_ix_i = 162$From the table,we have $\sum f_i = 20$ and $\sum f_ix_i = 162$.The mean $\bar{x}$ is given by $\bar{x} = \frac{\sum f_ix_i}{\sum f_i}$.Mean $= \frac{162}{20} = 8.1$.Therefore,the mean number of plants per house is $8.1$.The direct method is used here because the numerical values of $x_i$ and $f_i$ are small,making the calculations simple.

Daily income (in Rs.)	$100-120$	$120-140$	$140-160$	$160-180$	$180-200$
Number of workers	$12$	$14$	$8$	$6$	$10$

Similar Questions

If the mean of $12, 13, x, 17, 18$ and $20$ is $16,$ then $x = \ldots \ldots \ldots \ldots .$

The following distribution gives the daily income of $50$ workers of a factory.
Daily income (in Rs.) $100-120$ $120-140$ $140-160$ $160-180$ $180-200$
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Number of plants	$0-2$	$2-4$	$4-6$	$6-8$	$8-10$	$10-12$	$12-14$
Number of houses	$1$	$2$	$1$	$5$	$6$	$2$	$3$