Consider the following frequency distribution | vedclass

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(A) To find the median class,we first convert the discontinuous class intervals into continuous ones by adjusting the boundaries. The difference betwe

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$17.5$

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(A) To find the median class,we first convert the discontinuous class intervals into continuous ones by adjusting the boundaries. The difference between the upper limit of one class and the lower limit of the next is $1$. Thus,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit.						Class			Frequency $(f)$			Cumulative Frequency $(cf)$							$-0.5-5.5$			$13$			$13$							$5.5-11.5$			$10$			$23$							$11.5-17.5$			$15$			$38$							$17.5-23.5$			$8$			$46$							$23.5-29.5$			$11$			$57$			The total frequency $N = 57$. The median position is $N/2 = 57/2 = 28.5$. The cumulative frequency just greater than $28.5$ is $38$,which corresponds to the class interval $11.5-17.5$. Therefore,the median class is $11.5-17.5$. The upper limit of this class is $17.5$.

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Number of students	$2$	$4$	$7$	$6$	$1$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$10$	$12$	$13$	$16$	$9$

Class	$10-30$	$30-50$	$50-70$	$70-90$	$90-110$
Frequency	$17$	$f_{1}$	$32$	$f_{2}$	$19$

Consider the following frequency distribution:

Class $0-5$ $6-11$ $12-17$ $18-23$ $24-29$

Frequency $13$ $10$ $15$ $8$ $11$

The upper limit of the median class is

Similar Questions

Calculate the mean of the scores of $20$ students in a mathematics test:

Marks $10-20$ $20-30$ $30-40$ $40-50$ $50-60$

Number of students $2$ $4$ $7$ $6$ $1$

In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$,$f_{1}$ represents:

The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency $10$ $12$ $13$ $16$ $9$

For the frequency distribution given above,which of the following is false?

The mean of the following frequency distribution is $60$ and the total frequency is $120$. Find the missing frequencies $f_{1}$ and $f_{2}$.

Class $10-30$ $30-50$ $50-70$ $70-90$ $90-110$

Frequency $17$ $f_{1}$ $32$ $f_{2}$ $19$

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Class	$0-5$	$6-11$	$12-17$	$18-23$	$24-29$
Frequency	$13$	$10$	$15$	$8$	$11$

Consider the following frequency distribution: Class $0-5$ $6-11$ $12-17$ $18-23$ $24-29$ Frequency $13$ $10$ $15$ $8$ $11$ The upper limit of the median class is