Question

During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.

Answer 1

The given cumulative frequency distributions of less than type are

 

Weight (in kg upper class limits	Number of students (cumulative frequency)
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.

Here, n = 35

So  n/2 = 17.5

Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5

It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below

Weight (in kg)	Frequency (f)	Cumulative frequency
Less than 38	0	0
38 − 40	3 − 0 = 3	3
40 − 42	5 − 3 = 2	5
42 − 44	9 − 5 = 4	9
44 − 46	14 − 9 = 5	14
46 − 48	28 − 14 = 14	28
48 − 50	32 − 28 = 4	32
50 − 52	35 − 32 = 3	35
Total (n)	35

The cumulative frequency just greater than n/2 (35/2  = 17.5) is 28, belonging to class interval 46 − 48.

Median class = 46 − 48

Lower class limit (l) of median class = 46

Frequency (f) of median class = 14

Cumulative frequency (cf) of class preceding median class = 14

Class size (h) = 2

`"Median" = l + ((n/2-cf)/f)xxh`

= `46 + ((17.5-14)/14)xx2`

= 46+ 3.5/7

=  46.5

Therefore, median of this data is 46.5.

Hence, the value of median is verified.