question_answer

Following table shows marks (out of 100) of students in a class test:
Marks	No. of students
More than or equal to 0	80
More than or equal to 10	77
More than or equal to 20	72
More than or equal to 30	65
More than or equal to 40	55
More than or equal to 50	43
More than or equal to 60	28
More than or equal to 70	16
More than or equal to 80	10
More than or equal to 90	8
More than or equal to 100	0
Draw a ?more than type? ogive. From the curve, find the median. Also, check the value of the median by actual calculation.

Answer:

More than type	C.I.	No. of Students	Frequency	c.f.
More than or equal to 0	0 ? 10	80	3	3
More than or equal to 10	10 ? 20	77	5	8
More than or equal to 20	20 ? 30	72	7	15
More than or equal to 30	30 ? 40	65	10	25
More than or equal to 40	40 ? 50	55	12	37
More than or equal to 50	50 ? 60	43	15	52
More than or equal to 60	60 ? 70	28	12	64
More than or equal to 70	70 ? 80	16	06	70
More than or equal to 80	80 ? 90	10	02	72
More than or equal to 90	90 ? 100	8	08	80
More than or equal to 100	100 ? 110	0	00
Median will be 52
Median by actual calculation:
\[N=80\] (even)
\[=\frac{80}{2}\]
\[=40\]
So modal class will be \[5060\]
\[l=50,\,h=10,\,f=15,\]
\[c.f.=37\],
Median \[=l+\left[ h\times \frac{\left( \frac{N}{2}-c.f. \right)}{f} \right]\]
\[=50+\left[ 10\frac{(40-37)}{15} \right]\]
\[=50+10\times \frac{3}{15}\]
\[=50+2\]
\[=52\]              Hence Verified.