Paper 4: Fundamentals of Business Mathematics & Statistic

FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS I 5.27

Solution : The class intervals: are in discrete forms, so they are to be made in class boundaries at first

Table : Calculation of Median

Class boundaries f c.f

0.5–20.5 3 3

20.5–40.5 5 8

40.5–60.5 9 17

60.5-–80.5 3 20

80.5–100.5 2 22 (=N)

Median = value of^222 th term = values of 11th term ∴ Median class is (40.5 – 60.5)

Median = (^12) m^1
l l l
f

• − N c
  2

 − 

 = 40.5 +60.5 - 40.5 9 (11- 8)

= 40.5 +^209 ×^3 = 40.5 + 6.67 = 47.17 marks.

Calculation of median from open ends class intervals :
Since the first and last class intervals are not required in computing median, so in case of open end class-
intervals median is calculated by usual process.
For example, in the above example it the lower-limit of first class interval (i.e.0) and upper limit of last class
(i.e. 5) are not given question, there would be no difficulty to compute median.
In case of open end class-intervals, median is preferred than A.M. as average

Finding of missing frequency
The idea of finding missing frequency will be clear from the following example.

Example 33 : An incomplete frequency distributions given below :

Marks No. of students(f)

10–20 3

20–30 5

30–40 ––

40–50 3

50–60 1

It is given that median of the above distribution is 32.5 marks. Find the missing frequency.