Question

Consider the following frequency distribution:

Class	0 – 5	6 – 11	12 – 17	18 – 23	24 – 29
Frequency	13	10	15	8	11

Find the upper limit of the median class.

Answer 1

1. Convert to continuous classes

Since the given classes are discontinuous (0 – 5, 6 – 11, etc), we must convert them into a continuous frequency distribution by calculating an adjustment factor:

Adjustment factor = `("Lower limit of second class" - "Upper limit of first class")/2`

= `(6 - 5)/2`

= 0.5

Subtract 0.5 from each lower limit and add 0.5 to each upper limit to get the continuous classes. 

2. Form the cumulative frequency table

Now, let’s list the modified classes along with their frequencies (f) and cumulative frequencies (cf):

Discontinuous Class	Continuous Class	Frequency (f)	Cumulative frequency (cf)
0 – 5	–0.5 – 5.5	13	13
6 – 11	5.5 – 11.5	10	13 + 10 = 23
12 – 17	11.5 – 17.5	15	23 + 15 = 38
18 – 23	17.5 – 23.5	8	38 + 8 = 46
24 – 29	23.5 – 29.5	11	46 + 11 = 57

3. Determine the median class

Find the total frequency (N):

N = 57

`N/2 = 57/2 = 28.5`

The cumulative frequency that is just greater than or equal to 28.5 is 38. The corresponding continuous class interval is 11.5 – 17.5, making it our median class.

The upper limit of the median class is 17.5.