Question

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

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

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

Answer 1

To represent the data in the table graphically, we mark the upper limits of the class interval on x-axis and their corresponding cumulative frequency on y-axis choosing a convenient scale.

Now, let us plot the points corresponding to the ordered pair given by (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them by a freehand smooth curve

Thus, the curve obtained is the less than type ogive.

Now, locate n/2=35/2 =17.5 on the y- axis,

We draw a line from this point parallel to x-axis cutting the curve at a point. From this point, draw a perpendicular line to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data. Here it is 46.5. Let us make the following table in order to find median by using formula.

Weight (in kg)	No. of Students (frequency) (fi)	Cumulative frequency (cf)
36–38	0	0
38–40	3	3
40–42	2	5
42–44	4	9
44–46	5	14
46–48	14	28
48–50	4	32
50–52	3	35
Total	∑fi = 35

Hence, median is verified.