Draw an ogive for the following distribution. Determine the median graphically and verify your result by mathematical formula.
| Height (in cms.) | No. of students |
| 145 − 150 | 2 |
| 150 − 155 | 5 |
| 155 − 160 | 9 |
| 160 − 165 | 15 |
| 165 − 170 | 16 |
| 170 − 175 | 7 |
| 175 − 180 | 5 |
| 180 − 185 | 1 |
Solution
To draw a ogive curve, we construct the less than cumulative frequency table as given below:
| Height (in cms) | No. of students (f) |
Less than cumulative frequency (c.f.) |
| 145 – 150 | 2 | 2 |
| 150 – 155 | 5 | 7 |
| 155 – 160 | 9 | 16 |
| 160 – 165 | 15 | 31 |
| 165 – 170 | 16 | 47 |
| 170 – 175 | 7 | 54 |
| 175 – 180 | 5 | 59 |
| 180 – 185 | 1 | 60 |
| Total | 60 |
The points to be plotted for less than ogive are (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).
N = 60
∴ `"N"/2=60/2` = 30
∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis. From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis. Foot of the perpendicular gives the value of median.
∴ Median ≈ 164.67
Now, let us calculate the median from the mathematical formula.
∵ `"N"/2` = 30
The median lies in the class interval of 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16
Median = `"L"+"h"/"f"("N"/2-"c.f.")`
= `160 + 5/15 (30 - 16)`
= `160 + 1/3 xx 14`
= 160 + 4.67
= 164.67
