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प्रश्न
From the following frequency, prepare the ‘more than’ ogive.
| Score | Number of candidates |
| 400 – 450 | 20 |
| 450 – 500 | 35 |
| 500 – 550 | 40 |
| 550 – 600 | 32 |
| 600 – 650 | 24 |
| 650 – 700 | 27 |
| 700 – 750 | 18 |
| 750 – 800 | 34 |
| Total | 230 |
Also, find the median.
उत्तर
From the given table, we may prepare than ‘more than’ frequency table as shown below:
| Score | Number of candidates |
| More than 750 | 34 |
| More than 700 | 52 |
| More than 650 | 79 |
| More than 600 | 103 |
| More than 550 | 135 |
| More than 500 | 175 |
| More than 450 | 210 |
| More than 400 | 230 |
We plot the points A(750, 34), B(700,52),
C(650, 79), D(600, 103), E(550, 135), F(500, 175),
G(450, 210) and H(400, 230).
Join AB, BC, CD, DE, EF, FG, GH and HA with
a free hand to get the curve representing the
‘more than type’ series.
Here, N = 230
⇒ `N/2 = 115`
From P (0, 115), draw PQ meeting the curve at Q. Draw QM meeting at M.
Clearly, OM = 590 units
Hence, median = 590 units.
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| Production yield (in kg/ha) | 50 − 55 | 55 − 60 | 60 − 65 | 65 − 70 | 70 − 75 | 75 − 80 |
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Draw a ‘more than’ ogive for the data given below which gives the marks of 100 students.
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| No of Students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |
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| 140 – 144 | 3 |
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| 152 – 156 | 31 |
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| 160 – 164 | 64 |
| 164 – 168 | 75 |
| 168 – 172 | 82 |
| 172 – 176 | 86 |
| 176 – 180 | 34 |
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day:
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The marks obtained by 100 students of a class in an examination are given below.
| Mark | No. of Student |
| 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 'a less than' type cumulative frequency curves (ogive). Hence find the median.
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| Frequency | z | 5 | 6 | 1 | 2 |
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