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Question
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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Solution
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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