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प्रश्न
The distribution of heights (in cm) of 100 people is given below. Construct a histogram and the frequency polygon imposed on it.
| Height (in cm) |
125 − 135 | 136 − 146 | 147 − 157 | 158 − 168 | 169 − 179 | 180 − 190 | 191 − 201 |
| Frequency | 12 | 22 | 18 | 24 | 15 | 7 | 2 |
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उत्तर
The given distribution is discontinuous.
Converting into continuous distribution we have
Lower boundary = lower limit – `1/2` ...(gap between adjacent class interval)
= `125 - 1/2`
= 124.5
Upper boundary = Upper limit + `1/2` ...(gap between the adjacent class interval)
= `135 + 1/2`
= 135.5
∴ The new frequency table is
| Height (in cm) |
124.5 − 135.5 | 135.5 − 146.5 | 146.5 − 157.5 | 157.5 − 168.5 | 168.5 − 179.5 | 179.5 − 190.5 | 190.5 − 201.5 |
| Frequency | 12 | 22 | 18 | 24 | 15 | 7 | 2 |
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संबंधित प्रश्न
Draw a frequency polygon for the following grouped frequency distribution table.
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Age of the donor
(Yrs.) |
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| No. of blood doners | 38 | 46 | 35 | 24 | 15 | 12 |
The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.
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0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |
| No. of towns | 14 | 12 | 36 | 48 | 40 |
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