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(B) To find the variance,we first calculate the mid-interval values $(x)$ and the frequency $(f)$:Marks$x$$f$$f \cdot x$$f \cdot x^2$$1-3$$2$$40$$80$$

Vedclass · Accepted Answer

$4$

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(B) To find the variance,we first calculate the mid-interval values $(x)$ and the frequency $(f)$:Marks$x$$f$$f \cdot x$$f \cdot x^2$$1-3$$2$$40$$80$$160$$3-5$$4$$30$$120$$480$$5-7$$6$$20$$120$$720$$7-9$$8$$10$$80$$640$Total$100$$400$$2000$The mean $\bar{x} = \frac{\sum f \cdot x}{\sum f} = \frac{400}{100} = 4$.The variance $\sigma^2 = \frac{\sum f \cdot x^2}{\sum f} - (\bar{x})^2$.$\sigma^2 = \frac{2000}{100} - (4)^2 = 20 - 16 = 4$.

Class	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$1$	$3$	$4$	$2$

The variance of the following distribution is:
Marks $1-3$ $3-5$ $5-7$ $7-9$
Number of students $40$ $30$ $20$ $10$

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The variance of the following distribution is: Marks$1-3$$3-5$$5-7$$7-9$Number of students$40$$30$$20$$10$