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प्रश्न
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 |
उत्तर
(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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संबंधित प्रश्न
The marks obtained by 100 students of a class in an examination are given below:
Marks | Number of students |
0 – 5 | 2 |
5 – 10 | 5 |
10 – 15 | 6 |
15 – 20 | 8 |
20 – 25 | 10 |
25 – 30 | 25 |
30 – 35 | 20 |
35 – 40 | 18 |
40 – 45 | 4 |
45 – 50 | 2 |
Draw cumulative frequency curves by using (i) ‘less than’ series and (ii) ‘more than’ series.Hence, find the median.
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Number of Students | 2 | 7 | 8 | 30 | 12 | 1 |
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The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.
Class interval : | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
Frequency : | 4 | x | 5 | y | 1 |
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Class-interval: | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
Frequency: | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |
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C.I. | 0 - 5 | 6 - 11 | 12 - 17 | 18 - 23 | 24 - 29 |
f | 13 | 10 | 15 | 8 | 11 |
the upper limit of the median class is?
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Marks (out of 90) | Number of candidates |
More than or equal to 80 | 4 |
More than or equal to 70 | 6 |
More than or equal to 60 | 11 |
More than or equal to 50 | 17 |
More than or equal to 40 | 23 |
More than or equal to 30 | 27 |
More than or equal to 20 | 30 |
More than or equal to 10 | 32 |
More than or equal to 0 | 34 |
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Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Form: Less than type cumulative frequency distribution.