Q.Explain how the median of a continuous series can be located graphically using an ogive, and verify the graphic method against the algebraic median found in Worked Example 8.
Constructing the two ogives, using the Worked Example 8 data (class intervals 0–10 to 40–50, frequencies 5, 8, 15, 16, 6, N = 50):
- Less-than ogive. Plot the upper boundary of each class against its cumulative frequency, built from the lowest class upward: (10, 5), (20, 13), (30, 28), (40, 44), (50, 50). Joining these points gives a smoothly rising curve.
- More-than ogive. Plot the lower boundary of each class against the cumulative frequency of "this value or more", built from the highest class downward: (0, 50), (10, 45), (20, 37), (30, 22), (40, 6). Joining these points gives a smoothly falling curve.
Reading the median. The two curves intersect at exactly one point. Dropping a perpendicular from that point of intersection to the x-axis gives the median directly, without any interpolation arithmetic. For this data, the intersection occurs at x ≈ 28.
Why this must agree with the algebraic answer. Both curves are simply graphical plots of the same cumulative-frequency data used in the interpolation formula of §6 — the less-than curve reaches cf = 25 (that is, N/2) at exactly the x-value the interpolation formula computes, and the more-than curve reaches the complementary cf = 25 at the same x-value from the other direction. The two methods are two different ways of asking the identical question ("at what value does exactly half the data lie below?"), so they cannot legitimately disagree.
The ogive method locates the median at the intersection of the less-than and more-than cumulative-frequency curves, which for this data occurs at x ≈ 28 — matching the algebraic median of 28 found by the interpolation formula in Worked Example 8.
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