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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.
उत्तर
(i) From the given table, we may prepare the ‘less than’ frequency table as shown below:
Marks | Number of students |
Less than 5 | 2 |
Less than 10 | 7 |
Less than 15 | 13 |
Less than 20 | 21 |
Less than 25 | 31 |
Less than 30 | 56 |
Less than 35 | 76 |
Less than 40 | 94 |
Less than 45 | 98 |
Less than 50 | 100 |
We plot the points A(5, 2), B(10, 7), C(15, 13), D(20, 21), E(25, 31), F(30, 56), G(35, 76) and H(40, 94), I(45, 98) and J(50, 100).
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 0 | 100 |
More than 5 | 98 |
More than 10 | 93 |
More than 15 | 87 |
More than 20 | 79 |
More than 25 | 69 |
More than 30 | 44 |
More than 35 | 24 |
More than 40 | 6 |
More than 45 | 2 |
Now, on the same graph paper, we plot the points (0, 100), (5, 98), (10, 94), (15, 76), (20, 56), (25, 31), (30, 21), (35, 13), (40, 6) and (45, 2). Join 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 = 29.5
Hence, Median = 29.5
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Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 8 | 10 | 10 | 16 | 12 | 6 | 7 |
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Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
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