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(A) To find the median,we first calculate the cumulative frequency $(cf)$:| Class | Frequency $(f)$ | Cumulative Frequency $(cf)$ || :--- | :--- | :--

Vedclass · Accepted Answer

$153.79$

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(A) To find the median,we first calculate the cumulative frequency $(cf)$:| Class | Frequency $(f)$ | Cumulative Frequency $(cf)$ || :--- | :--- | :--- || $110-120$ | $6$ | $6$ || $120-130$ | $25$ | $31$ || $130-140$ | $48$ | $79$ || $140-150$ | $72$ | $151$ || $150-160$ | $116$ | $267$ || $160-170$ | $60$ | $327$ || $170-180$ | $38$ | $365$ || $180-190$ | $22$ | $387$ || $190-200$ | $3$ | $390$ |Total frequency $N = 390$. So,$N/2 = 390/2 = 195$.The cumulative frequency just greater than $195$ is $267$,which corresponds to the class interval $150-160$.Thus,the median class is $150-160$.Here,lower limit $l = 150$,frequency of median class $f = 116$,cumulative frequency of the class preceding the median class $cf = 151$,and class size $h = 10$.Using the median formula: $	ext{Median} = l + \left( \frac{N/2 - cf}{f} ight) 	imes h$$	ext{Median} = 150 + \left( \frac{195 - 151}{116} ight) 	imes 10$$	ext{Median} = 150 + \left( \frac{44}{116} ight) 	imes 10 = 150 + \frac{440}{116} \approx 150 + 3.79 = 153.79$.

Class	$21-25$	$26-30$	$31-35$	$36-40$	$41-45$	$46-50$	$51-55$
Frequency	$18$	$32$	$30$	$40$	$25$	$15$	$40$

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Frequency	$5$	$x$	$20$	$y$	$2$

Find the median of the following frequency distribution:

Class Frequency $(f)$

$110-120$ $6$

$120-130$ $25$

$130-140$ $48$

$140-150$ $72$

$150-160$ $116$

$160-170$ $60$

$170-180$ $38$

$180-190$ $22$

$190-200$ $3$

Similar Questions

Find the mean of the following frequency distribution:

Class $21-25$ $26-30$ $31-35$ $36-40$ $41-45$ $46-50$ $51-55$

Frequency $18$ $32$ $30$ $40$ $25$ $15$ $40$

(in $.675$)

The median of the following frequency distribution is $50$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.

Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$

Frequency $5$ $x$ $20$ $y$ $2$

.......... is not a measure of central tendency.

For some given data,if $Z = 95$ and $\bar{x} = 98,$ then $M = \ldots \ldots \ldots$

While computing the mean of grouped data,we assume that the frequencies are

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Class	Frequency $(f)$
$110-120$	$6$
$120-130$	$25$
$130-140$	$48$
$140-150$	$72$
$150-160$	$116$
$160-170$	$60$
$170-180$	$38$
$180-190$	$22$
$190-200$	$3$

Find the median of the following frequency distribution: Class Frequency $(f)$ $110-120$ $6$ $120-130$ $25$ $130-140$ $48$ $140-150$ $72$ $150-160$ $116$ $160-170$ $60$ $170-180$ $38$ $180-190$ $22$ $190-200$ $3$