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प्रश्न
The following distribution represents the height of 160 students of a school.
| Height (in cm) | No. of Students |
| 140 – 145 | 12 |
| 145 – 150 | 20 |
| 150 – 155 | 30 |
| 155 – 160 | 38 |
| 160 – 165 | 24 |
| 165 – 170 | 16 |
| 170 – 175 | 12 |
| 175 – 180 | 8 |
Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:
- The median height.
- The interquartile range.
- The number of students whose height is above 172 cm.
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उत्तर
| Height (in cm) |
No. of students |
Cumulative frequency |
| 140 – 145 | 12 | 12 |
| 145 – 150 | 20 | 32 |
| 150 – 155 | 30 | 62 |
| 155 – 160 | 38 | 100 |
| 160 – 165 | 24 | 124 |
| 165 – 170 | 16 | 140 |
| 170 – 175 | 12 | 152 |
| 175 – 180 | 8 | 160 |
| N = 160 |
Taking height of student along x-axis and cumulative frequency along y-axis we will draw an ogive.

i. ∴ Median = `160/2` = 80th term
Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5.
ii. Since, number of terms = 160
Lower quartile (Q1) = `(160/4)` = 40th term = 152
Upper quartile (Q3) = `((3 xx 160)/4)` = 120th term = 164
Inner Quartile range = Q3 – Q1
= 164 – 152
= 12
iii. Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145.
The number of students whose height is above 172 cm = 160 – 144 = 16
