Question

how to find median for discontinuous frequency

Answer 1

search Google Engine you will see your answer.


I hope it helps

Answer 2

Example: The following distribution represents the number  of  minutes  spent per week by a group  of teenagers in going to the movies. Find the median number of minutes spent per by the teenagers in going to the movies.

Number of  minutes  per week

Number of teenagers

0-99

26

100-199

32

200-299

65

300-399

75

400-499

60

500-599

42

Solution:

Let us convert  the class intervals given, to class boundaries and construct the less than type cumulative frequency distribution.

Number of  minutes  per weekClass BoundariesNumber of teenagers (Frequency)

Cumulative Frequency

(less than type)

0-99

0-99.526

26

100-199

99.5-199.532

58

200-299

199.5-299.565

123

300-399

299.5-399.575

198

400-499

399.5-499.560

258

500-599

499.5-599.542

300

Here, 

Here, the cumulative frequency just greater than or equal  to 150 is 198.

 and

 is the less than type cumulative frequency corresponding to the class boundary 399.5



 the median class is the class for which upper class boundary is 

In other words, 299.5-399.5 is the median class, i.e. the class containing the median value.

 using the formula for  median we have,

Median or  

where,  (lower class boundary of the median class),

 (total frequency),

 ( less than type cumulative frequency corresponding to ),

 (frequency of  the median class),

and  (class width of the median class).



or, Median (

So, the median number of minutes spent per week  by this group of 300 teenagers in going to the movies is 335.5, i.e. there are 150 teenagers for whom the number of minutes spent per week in going to the movies is less than 335.5 and there are another 150 teenagers for whom the number of minutes spent per week in going to the movies is greater than 335.5.