Question

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

Answer 1

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