Question

The following table gives the marks obtained by $300$ students of a management course. Find the median of the distribution.

Marks obtained	No. of students
$0-15$	$26$
$15-30$	$34$
$30-45$	$64$
$45-60$	$76$
$60-75$	$60$
$75-90$	$60$
$90-100$	$10$

• A. $62.96$ marks
• B. $62.13$ marks
• C. $50.13$ marks
• D. $50.96$ marks

Answer 1

The correct option is C $50.13$ marks

Median is the middle most value of a given series that represents the whole class of the series. For a group data,

Median = L + [{(n/2) – B}/G] × w where

• Lis the lower class boundary of the group containing the median
  
• nis the total number of values
  
• Bis the cumulative frequency of the groups before the median group
  
• Gis the frequency of the median group
  
• wis the group width
  

Since the median is the middle value, which in this case is 165th one , which is in the 45-60 group. Therefore , 45-60 is the median group so

• L= 45
  
• n= 330
  
• B= 26 + 34 + 64 = 124
  
• G= 76
  
• w= 15
  

Median= 45 + [{(330/2) – 124}/76] × 15

• = 45 + 8.09
  
• = 53.09