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प्रश्न
The distribution of heights (in cm) of 96 children is given below. Construct a histogram and a frequency polygon on the same axes.
| Height (in cm): | 124 to 128 |
128 to 132 |
132 to 136 |
136 to 140 |
140 to 144 |
144 to 148 |
148 to 152 |
152 to 156 |
156 to 160 |
160 to 164 |
| No. of Children: | 5 | 8 | 17 | 24 | 16 | 12 | 6 | 4 | 3 | 1 |
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उत्तर
To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights.
To draw the frequency polygon of the given data using histogram, obtain the mid-points of the upper horizontal side of each rectangle and then join these mid-points of the adjacent rectangles of the histogram by line segments. Obtain the mid-points of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Complete the polygon by joining the mid-points of first and last class-intervals to the mid-points of imagined class-intervals adjacent to them. Let us take one vertical division is equal to 4.
The heights of the different rectangles are as following
1. The height of the rectangle corresponding to the class-interval 124-128 is `5/4 = 1.25` big divisions.
2. The height of the rectangle corresponding to the class-interval 128-132 is ` 8/4`=2` big divisions.
3. The height of the rectangle corresponding to the class-interval 132-136 is `17/4 = 4.25` big divisions.
4. The height of the rectangle corresponding to the class-interval 136-140 is `24/4 = 6` big divisions.
5. The height of the rectangle corresponding to the class-interval 140-144 is `16/4 = 4` big divisions.
6. The height of the rectangle corresponding to the class-interval 144-148 is `12/4 = 3` big divisions.
7. The height of the rectangle corresponding to the class-interval 148-152 is `6/4 = 1.5` big divisions.
8. The height of the rectangle corresponding to the class-interval 152-156 is `4/4 =1 ` big divisions.
9. The height of the rectangle corresponding to the class-interval 156-160 is `3/4 = 0.75` big divisions.
10. The height of the rectangle corresponding to the class-interval 160-164 is `1/4 = 0.25 ` big divisions.
The histogram and frequency polygon of the given data is the following:
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संबंधित प्रश्न
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| 1. | Reproductive health conditions | 31.8 |
| 2. | Neuropsychiatric conditions | 25.4 |
| 3. | Injuries | 12.4 |
| 4. | Cardiovascular conditions | 4.3 |
| 5. | Respiratory conditions | 4.1 |
| 6. | Other causes | 22.0 |
- Represent the information given above graphically.
- Which condition is the major cause of women’s ill health and death worldwide?
- Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
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| Section | Number of girls per thousand boys |
| Scheduled Caste (SC) | 940 |
| Scheduled Tribe (ST) | 970 |
| Non SC/ST | 920 |
| Backward districts | 950 |
| Non-backward districts | 920 |
| Rural | 930 |
| Urban | 910 |
- Represent the information above by a bar graph.
- In the classroom discuss what conclusions can be arrived at from the graph.
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| 300 - 400 | 14 |
| 400 - 500 | 56 |
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- Represent the given information with the help of a histogram.
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Daily wages (in Rs.) |
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225 - 275 |
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325 - 375 |
|
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4 |
20 |
22 |
10 |
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Draw a frequency polygon for this distribution.
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| 85 | 60 | 35 | 50 | 70 |
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| Marks as percent | 60 | 45 | 42 | 48 | 75 |
Represent the above data by drawing a suitable bar graph.
