The lengths of 40 leaves of a plant are measu | vedclass

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(C) The given data does not have continuous class intervals. The difference between the upper limit of one class and the lower limit of the next class

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$146.75$

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(C) The given data does not have continuous class intervals. The difference between the upper limit of one class and the lower limit of the next class is $127 - 126 = 1$. Therefore,we subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of each class to make them continuous.Continuous class intervals and cumulative frequencies are as follows:Length (in mm)Frequency $(f_i)$Cumulative Frequency $(cf)$$117.5-126.5$$3$$3$$126.5-135.5$$5$$8$$135.5-144.5$$9$$17$$144.5-153.5$$12$$29$$153.5-162.5$$5$$34$$162.5-171.5$$4$$38$$171.5-180.5$$2$$40$Here,$n = 40$,so $\frac{n}{2} = 20$. The cumulative frequency just greater than $20$ is $29$,which corresponds to the class interval $144.5-153.5$.Median class $= 144.5-153.5$,lower limit $(l)$ $= 144.5$,class size $(h)$ $= 9$,frequency $(f)$ $= 12$,and cumulative frequency of the preceding class $(cf)$ $= 17$.Using the median formula: $	ext{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} ight) 	imes h$$	ext{Median} = 144.5 + \left( \frac{20 - 17}{12} ight) 	imes 9 = 144.5 + \left( \frac{3}{12} ight) 	imes 9 = 144.5 + 2.25 = 146.75$.Thus,the median length of the leaves is $146.75 	ext{ mm}$.

Monthly income range (in $Rs.$)	Number of families
Income more than $10000$	$100$
Income more than $13000$	$85$
Income more than $16000$	$69$
Income more than $19000$	$50$
Income more than $22000$	$33$
Income more than $25000$	$15$

Class	$12.5-17.5$	$17.5-22.5$	$22.5-27.5$	$27.5-32.5$	$32.5-37.5$
Frequency	$2$	$22$	$19$	$14$	$13$

Similar Questions

Cumulative frequency curve is also known as ........

Construct the cumulative frequency distribution of the following distribution:

Class $12.5-17.5$ $17.5-22.5$ $22.5-27.5$ $27.5-32.5$ $32.5-37.5$

Frequency $2$ $22$ $19$ $14$ $13$

In the formula $M = l + \frac{(\frac{n}{2} - cf)}{f} \times c$ for the median,$cf = \ldots \ldots \ldots$

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its

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Length (in mm)	Number of leaves
$118-126$	$3$
$127-135$	$5$
$136-144$	$9$
$145-153$	$12$
$154-162$	$5$
$163-171$	$4$
$172-180$	$2$

The lengths of $40$ leaves of a plant are measured correct to the nearest millimetre,and the data obtained is represented in the following table:Length (in mm)Number of leaves$118-126$$3$$127-135$$5$$136-144$$9$$145-153$$12$$154-162$$5$$163-171$$4$$172-180$$2$Find the median length of the leaves.