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(B) Arranging the data in ascending order of magnitude,we obtain:      Height (in $cm$)    $150$    $152$    $154$    $155$    $156$    $160$    $161$

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$155$

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(B) Arranging the data in ascending order of magnitude,we obtain:      Height (in $cm$)    $150$    $152$    $154$    $155$    $156$    $160$    $161$        Number of students    $8$    $4$    $3$    $7$    $3$    $12$    $4$        Cumulative frequency    $8$    $12$    $15$    $22$    $25$    $37$    $41$  Here,the total number of items $N = 41$,which is an odd number.Hence,the median is the $\left(\frac{N + 1}{2}\right)^{th}$ item.Median = $\left(\frac{41 + 1}{2}\right)^{th} = 21^{st}$ item.From the cumulative frequency table,we find that the $21^{st}$ item corresponds to a height of $155$ $cm$ (since all items from $16^{th}$ to $22^{nd}$ are equal to $155$).Therefore,the median is $155$ $cm$.

Class	Frequency
$10-20$	$180$
$20-30$	$82$
$30-40$	$34$
$40-50$	$180$
$50-60$	$136$
$60-70$	$23$
$70-80$	$50$

The following data gives the distribution of height of students. The median of the distribution is:

Height (in $cm$) $160$ $150$ $152$ $161$ $156$ $154$ $155$

No of students $12$ $8$ $4$ $4$ $3$ $3$ $7$

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The median of a set of $9$ distinct observations is $20.5$. If the four largest observations of the set are increased by $2$,what will be the median of the new set?

Find the median for the following distribution:
Class Frequency
$10-20$ $180$
$20-30$ $82$
$30-40$ $34$
$40-50$ $180$
$50-60$ $136$
$60-70$ $23$
$70-80$ $50$

The arithmetic mean of a set of observations is $\overline{X}$. If each observation is divided by $\alpha$ and then increased by $10$,then the mean of the new series is

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Height (in $cm$)	$160$	$150$	$152$	$161$	$156$	$154$	$155$
No of students	$12$	$8$	$4$	$4$	$3$	$3$	$7$

The following data gives the distribution of height of students. The median of the distribution is: Height (in $cm$) $160$ $150$ $152$ $161$ $156$ $154$ $155$ No of students $12$ $8$ $4$ $4$ $3$ $3$ $7$