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Question
From the following data, draw the two types of cumulative frequency curves and determine the median:
| Marks | Frequency |
| 140 – 144 | 3 |
| 144 – 148 | 9 |
| 148 – 152 | 24 |
| 152 – 156 | 31 |
| 156 – 160 | 42 |
| 160 – 164 | 64 |
| 164 – 168 | 75 |
| 168 – 172 | 82 |
| 172 – 176 | 86 |
| 176 – 180 | 34 |
Solution
(i) Less than series:
| Marks | Number of students |
| Less than 144 | 3 |
| Less than 148 | 12 |
| Less than 152 | 36 |
| Less than 156 | 67 |
| Less than 160 | 109 |
| Less than 164 | 173 |
| Less than 168 | 248 |
| Less than 172 | 230 |
| Less than 176 | 416 |
| Less than 180 | 450 |
We plot the points A(144, 3), B(148, 12), C(152, 36), D(156, 67), E(160, 109), F(164, 173), G(168, 248) and H(172, 330), I(176, 416) and J(180, 450).
Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.
(ii) More than series:
| Marks | Number of students |
| More than 140 | 450 |
| More than 144 | 447 |
| More than 148 | 438 |
| More than 152 | 414 |
| More than 156 | 383 |
| More than 160 | 341 |
| More than 164 | 277 |
| More than 168 | 202 |
| More than 172 | 120 |
| More than 176 | 34 |
Now, on the same graph paper, we plot the points A1(140, 450), B1(144, 447), C1(148, 438), D1(152, 414), E1(156, 383), F1(160, 277), H1(168, 202), I1(172, 120) and J1(176, 34).
Join A1B1, B1C1, C1D1, D1E1, E1F1, F1G1, G1H1, H1I1 and I1J1 with a free hand to get the ‘more than type’ series.
The two curves intersect at point L. Draw LM ⊥ OX cutting the x-axis at M. Clearly, M = 166cm
Hence, median = 166cm
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