Estimate Median, Quartiles from Ogive

For a frequency distribution, the median and quartiles can be obtained by drawing the ogive of the distribution. Follow these steps.

Step I: Change the frequency distribution into a continuous distribution by taking overlapping intervals. Let N be the total frequency.

Step II: Construct a cumulative-frequency table for the distribution and draw the ogive accordingly by using proper scales of representation.

Step III: For median (i) If N is odd, find $$\frac{N + 1}{2}$$, and locate the point F on the y-axis which represents the cumulative frequency $$\frac{N + 1}{2}$$.

(ii) If N is even, find the mean A of $$\frac{N}{2}$$ and $$\frac{N}{2}$$ + 1, which is given by A = $$\frac{1}{2}$${$$\frac{N}{2}$$ + ($$\frac{N}{2}$$ + 1)}. Locate the point F on the y-axis, which represents the cumulative frequency A.

For lower quartile: Find the integer c just greater than $$\frac{N}{4}$$. Locate the point F on the y-axis, which represents the cumulative frequency c.

For upper quartile: Find the integer c just greater than $$\frac{3N}{4}$$. Locate the point F on the y-axis, which represents the cumulative frequency c.

Step IV: Draw a line FD parallel to the x-axis to cut the ogive at C.

Step V: Draw a line CM perpendicular to the x-axis (class-interval axis) to cut the ogive at M. The variate represented by M is the median or lower quartile or upper quartile as the case may be.

Solved Problems on Estimate Median, Quartiles from Ogive:

1. Estimate the median, lower quartile and upper quartile for the following distribution.

Class Interval

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

Frequency

5

3

10

6

4

2

Solution:

Here, the distribution is continuous and total frequency = 30.

For constructing the ogive (step II), the following cumulative-frequency table is constructed.

Class Interval

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

Frequency

5

8

18

24

28

30

Take the following scales:

On the x-axis (class-interval axis), 1 cm = size 10.

On the y-axis (cumulative –frequency axis), 2 mm = frequency 1 (i.e., frequency of 1 is denoted by 2 mm).

Now, plot the pojnts (10, 5), (20, 8), (30, 18), (40, 24), (50, 28), (60, 30), and join them by a smooth curve to get the ogive.

Here, N = 30 = even. So, the mean of $$\frac{N}{2}$$ and $$\frac{N}{2}$$ + 1, i.e., the mean of 15 and 16, is 15.5. The point F on the y-axis represents the cumulative frequency 15.5. FC ∥ x-axis is drawn to cut the ogive at C. CM ⊥ x-axis is drawn to cut at M. The point M represents the median. Now, the point M represents the variate 28 on the x-axis.

So, the median is 28.

Now, $$\frac{N}{4}$$ = $$\frac{30}{4}$$ = 7.5. The integer just greater than 7.5 is 8. The point F1 on the y-axis represents the cumulative frequency 8. F1C1 ∥ x-axis is drawn to cut the ogive at C1. C1Q1 ⊥ x-axis is drawn to cut the ogive at Q1. The point Q1 represents the lower quartile. Now, the point Q1 represents the variate 20. So, the lower quartile is 20.

Next, $$\frac{3N}{4}$$ = $$\frac{3 × 30}{4}$$ = 22.5. The integer just greater than 22.5 is 23. The point F2 on the y-axis represents the cumulative frequency 23. F2C2 ∥ x-axis is drawn to cut the ogive at C2. C2Q2 ⊥ x-axis is drawn to cut the ogive at Q2. The point Q2 represents the upperr quartile. Now, the point Q2 represents the variate 38. So, the upper quartile is 38.

Note: Thse estimates are generally rough (that is, with marginal error) because the drawing of an ogive is never perfect.