Compute the median from the following table: | vedclass

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(A) To find the median,we first construct the cumulative frequency table:      Marks obtained    Frequency $(f)$    Cumulative Frequency $(cf)$

Vedclass · Accepted Answer

$36.55$

Vedclass · Answer

(A) To find the median,we first construct the cumulative frequency table:      Marks obtained    Frequency $(f)$    Cumulative Frequency $(cf)$        $0-10$    $2$    $2$        $10-20$    $18$    $20$        $20-30$    $30$    $50$        $30-40$    $45$    $95$        $40-50$    $35$    $130$        $50-60$    $20$    $150$        $60-70$    $6$    $156$        $70-80$    $3$    $159$  Here,$N = \sum f = 159$.We calculate $\frac{N}{2} = \frac{159}{2} = 79.5$.The cumulative frequency just greater than $79.5$ is $95$,which corresponds to the class interval $30-40$.Thus,the median class is $30-40$.Here,$l = 30$,$f = 45$,$cf = 50$,and $h = 10$.Using the median formula: $	ext{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} ight) 	imes h$$	ext{Median} = 30 + \left( \frac{79.5 - 50}{45} ight) 	imes 10$$	ext{Median} = 30 + \left( \frac{29.5}{45} ight) 	imes 10 = 30 + \frac{295}{45} = 30 + 6.555... \approx 36.56$.Rounding to two decimal places,we get $36.56$. Since $36.55$ is the closest option provided,the correct answer is $A$.

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

Compute the median from the following table:

Marks obtained No. of students

$0-10$ $2$

$10-20$ $18$

$20-30$ $30$

$30-40$ $45$

$40-50$ $35$

$50-60$ $20$

$60-70$ $6$

$70-80$ $3$

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

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Marks obtained	No. of students
$0-10$	$2$
$10-20$	$18$
$20-30$	$30$
$30-40$	$45$
$40-50$	$35$
$50-60$	$20$
$60-70$	$6$
$70-80$	$3$

Compute the median from the following table: Marks obtained No. of students $0-10$ $2$ $10-20$ $18$ $20-30$ $30$ $30-40$ $45$ $40-50$ $35$ $50-60$ $20$ $60-70$ $6$ $70-80$ $3$