Draw ogive for the Following distribution and hence find graphically the limits of weight of middle 50% fishes.
| Weight of fishes (in gms) | 800 – 890 | 900 – 990 | 1000 – 1090 | 1100 – 1190 | 1200 – 1290 | 1300 –1390 | 1400 – 1490 |
| No. of fishes | 8 | 16 | 20 | 25 | 40 | 6 | 5 |
Solution
Since the given data is not continuous, we have to convert it in the continuous form by subtracting 5 from the lower limit and adding 5 to the upper limit of every class interval. To draw a ogive curve, we construct the less than cumulative frequency table as given below:
| Weight of fishes (in gms) |
No. of fishes (f) |
Less than cumulative frequency (c.f.) |
| 795 – 895 | 8 | 8 |
| 895 – 995 | 16 | 24 |
| 995 – 1095 | 20 | 44 |
| 1095 – 1195 | 25 | 69 |
| 1195 – 1295 | 40 | 109 |
| 1295 – 1395 | 6 | 115 |
| 1395 – 1496 | 5 | 120 |
| Total | 120 |
Points to be plotted are (895, 8), (995, 24),(1095, 44),(1195, 69),(1295, 109), (1395, 115), (1495, 120).
N = 120
For Q1 and Q3 we have to consider `"N"/4=120/4` = 30, `(3"N")/4=(3xx120)/4` = 90
For finding Q1 and Q3 we consider the values 30 and 90 on the Y-axis. From these points, we draw the lines which are parallel to X-axis. From the points where these lines intersect the less than ogive, we draw perpendicular on X-axis. The feet of perpendiculars represent the values of Q1 and Q2.
∴ Q1 ≈ 1025 and Q3 ≈ 1248
∴ The limits of weight of middle 50% fishes lie between 1025 to 1248.
