From the following distribution, determine median graphically.
| Daily wages (in ₹) | No. of employees |
| Above 300 | 520 |
| Above 400 | 470 |
| Above 500 | 399 |
| Above 600 | 210 |
| Above 700 | 105 |
| Above 800 | 45 |
| Above 900 | 7 |
Solution 1
The given ‘more than cumulative frequency’ table is,
| Daily Wages (in Rs.) |
No. of employee (m.c.f) |
| Above 300 | 520 |
| Above 400 | 470 |
| Above 500 | 399 |
| Above 600 | 210 |
| Above 700 | 105 |
| Above 800 | 45 |
| Above 900 | 7 |
The ‘more than ogive curve’ is plotted using the given lower limit of the class interval (x) against c.f. (y)
For median draw a line parallel to X-axis at Frequency `"N"/(2) = (520)/(2)` = 260
Solution 2
To draw a ogive curve, we construct the less than and more than cumulative frequency table as given below:
| Daily wages (in ₹) | No. of employees (f) |
Less than cumulative frequency (c.f.) |
More than cumulative frequency (c.f.) |
| 300 − 400 | 50 | 50 | 520 |
| 400 − 500 | 71 | 121 | 470 |
| 500 − 600 | 189 | 310 | 399 |
| 600 − 700 | 105 | 415 | 210 |
| 700 − 800 | 60 | 475 | 105 |
| 800 − 900 | 38 | 513 | 45 |
| 900 − 1000 | 7 | 520 | 7 |
| Total | 520 |
The points to be plotted for less than ogive are (400, 50), (500, 121), (600, 310), (700, 415), (800, 475), (900, 513) and (1000, 520) and that for more than ogive are (300, 520), (400, 470), (500, 399), (600, 210), (700, 105), (800, 45), (900, 7).
From the point of intersection of two ogives, we draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.
∴ Median ≈ 574
