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 F_{1} on the y-axis represents the cumulative frequency 8. F_{1}C_{1} ∥ x-axis is drawn to cut the ogive at C_{1}. C_{1}Q_{1} ⊥ x-axis is drawn to cut the ogive at Q_{1}. The point Q_{1} represents the lower quartile. Now, the point Q_{1} 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 F_{2} on the y-axis represents the cumulative frequency 23. F_{2}C_{2} ∥ x-axis is drawn to cut the ogive at C_{2}. C_{2}Q_{2} ⊥ x-axis is drawn to cut the ogive at Q_{2}. The point Q_{2} represents the upperr quartile. Now, the point Q_{2} 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.
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