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प्रश्न
The lengths of 62 leaves of a plant are measured in millimetres and the data is represented in the following table:
| Length (in mm) | Number of leaves |
| 118 – 126 | 8 |
| 127 – 135 | 10 |
| 136 – 144 | 12 |
| 145 – 153 | 17 |
| 154 – 162 | 7 |
| 163 – 171 | 5 |
| 172 – 180 | 3 |
Draw a histogram to represent the data above.
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उत्तर
The given frequency distribution is in inclusive form.
So, first we convert it into exclusive form.
Now, adjusting factor = `(127 - 126)/2 = 1/2 = 0.5`
So, we subtract 0.5 from each lower limit and add 0.5 to each upper limit.
The table for continuous grouped frequency distribution is given below:
| Length (in mm) | Number of leaves |
| 117.5 – 126.5 | 8 |
| 126.5 – 135.5 | 10 |
| 135.5 – 144.5 | 12 |
| 144.5 – 153.5 | 17 |
| 153.5 – 162.5 | 7 |
| 162.5 – 171.5 | 5 |
| 171.5 – 180.5 | 3 |
The table for continuous grouped frequency distribution is given below
Thus, the given data becomes in exclusive form.
Along the horizontal axis, we represent the class intervals of length on some suitable scale. The corresponding frequencies of number of leaves are represented along the Y-axis on a suitable scale.
Since, the given intervals start with 117.5 – 126.5. It means that, there is some break (vw) indicated near the origin to signify the graph is drawn with a scale beginning at 117.5.
A histogram of the given distribution is given below:

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संबंधित प्रश्न
Read the bar graph shown in Fig. 23.10 and answer the following questions
(i) What is the information given by the bar graph?

(ii) What was the number of commercial banks in 1977?
(iii) What is the ratio of the number of commercial banks in 1969 to that in 1980?
(iv) State whether true or false:
The number of commercial banks in 1983 is less than double the number of commercial banks in 1969.
The following bar graph shows the results of an annual examination in a secondary school. Read the bar graph and choose the correct alternative in each of the following:

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(a) VI, VIII (b) VI, IX (c) VIII, IX (d) VIII, X
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(a) VII (b) X (c) IX (d) VIII
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| Qualifications: | MCA (Master in Computer applications) |
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Construct a histogram for the following data:
| Monthly School fee (in Rs): |
30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 | 210-240 |
| No of Schools | 5 | 12 | 14 | 18 | 10 | 9 | 4 |
Draw, in the same diagram, a histogram and a frequency polygon to represent the following data which shows the monthly cost of living index of a city in a period of 2 years:
| Cost of living index: |
440-460 | 460-480 | 480-500 | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
| No. of months: | 2 | 4 | 3 | 5 | 3 | 2 | 1 | 4 |
In a frequency distribution, ogives are graphical representation of
Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram
|
C.I |
10 - 30 |
30 - 50 |
50 - 70 | 70 - 90 | 90 - 110 | 110 - 130 | 130 - 150 |
| ƒ | 4 | 7 | 5 | 9 | 5 | 6 | 4 |
Construct a frequency polygon for the following distribution:
| Class-intervals | 0-4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 | 20 - 24 |
| Frequency | 4 | 7 | 10 | 15 | 11 | 6 |
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46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44
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Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way:
| Class interval (km/h) | Frequency |
| 30 – 40 | 3 |
| 40 – 50 | 6 |
| 50 – 60 | 25 |
| 60 – 70 | 65 |
| 70 – 80 | 50 |
| 80 – 90 | 28 |
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Draw the frequency polygon representing the above data without drawing the histogram.
