Find the median of the following frequency di | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$153.79$

Vedclass · Answer

(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$.

Marks	$30-35$	$35-40$	$40-45$	$45-50$	$50-55$	$55-60$	$60-65$
Frequency	$14$	$16$	$18$	$23$	$18$	$8$	$3$

Class (seconds)	$13.8-14.0$	$14.0-14.2$	$14.2-14.4$	$14.4-14.6$	$14.6-14.8$	$14.8-15.0$
Frequency	$2$	$4$	$5$	$71$	$48$	$20$

Class	$20-25$	$25-30$	$30-35$	$35-40$	$40-45$	$45-50$	$50-55$
Frequency	$2$	$5$	$8$	$10$	$7$	$10$	$3$

Class interval	Frequency
$0-100$	$2$
$100-200$	$5$
$200-300$	$x$
$300-400$	$12$
$400-500$	$17$
$500-600$	$20$
$600-700$	$y$
$700-800$	$9$
$800-900$	$7$
$900-1000$	$4$

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

The percentage of marks obtained by $100$ students in an examination are given below:

Marks $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ $60-65$

Frequency $14$ $16$ $18$ $23$ $18$ $8$ $3$

Determine the median percentage of marks.

The times,in seconds,taken by $150$ athletes to run a $110\, m$ hurdle race are tabulated below:

Class (seconds) $13.8-14.0$ $14.0-14.2$ $14.2-14.4$ $14.4-14.6$ $14.6-14.8$ $14.8-15.0$

Frequency $2$ $4$ $5$ $71$ $48$ $20$

The number of athletes who completed the race in less than $14.6\, s$ is:

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

The median class in the following frequency distribution is ..........

Class $20-25$ $25-30$ $30-35$ $35-40$ $40-45$ $45-50$ $50-55$

Frequency $2$ $5$ $8$ $10$ $7$ $10$ $3$

The median of the following data is $525$. Find the values of $x$ and $y$,if the total frequency is $100$.
Class interval Frequency
$0-100$ $2$
$100-200$ $5$
$200-300$ $x$
$300-400$ $12$
$400-500$ $17$
$500-600$ $20$
$600-700$ $y$
$700-800$ $9$
$800-900$ $7$
$900-1000$ $4$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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