Question

Estimate the median, the lower quartile and the upper quartile of the following frequency distribution by drawing an ogive: 

Marks(more than)	90	80	70	60	50	40	30	20	10	0
No. of students	6	13	22	34	48	60	70	78	80	80

Answer 1

Given data is cumulative data , so draw the ogive as it is .

Marks (more than)	No. of students (f)
0	80
10	80
20	78
30	70
40	60
50	48
60	34
70	22
80	13
90	6

Take a graph paper and draw both the axes.

On the x-axis , take a scale of 1 cm = 10 to represent the marks (more than).

On the y - axis , take a scale of 1cm = 10 to represent the no. of students.

Now, plot the points (0,80) , (10,80) , (20,78) , (30,70) , (40,60) , (50,48) , (60,34) , (70,22) ,(80,13) , (90,6).

Join them by a smooth curve to get the ogive.

No. of terms = 80

∴ Median = ${\displaystyle \frac{40+41}{2}}$ = 40.5^th term

Through mark of 40.5 on y-axis draw a line parallel to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis which meets it at B.

The value of B is the median which is 55.

Lower Quartile (Q₁) = ${\displaystyle \frac{n}{4}=\frac{80}{4}}$ = 20^th term

Through mark of 20 on y-axis draw a line parallel to x-axis which meets the curve at P. From P , draw a perpendicular to x-axis which meets it at Q.

The value of Q is the lower quartile which is 71.

Upper Quartile (Q₃) = ${\displaystyle \frac{n\times 3}{4}=\frac{80\times 3}{4}}$ = 60^th term

Through mark of 60 on y-axis draw a line parallel to x-axis which meets the curve at R. From R, draw a perpendicular to x-axis which meets it at S.

The value of S is the upper Quartile which is 40.