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 Q_{1 }and Q_{3} we have to consider `"N"/4=120/4` = 30, `(3"N")/4=(3xx120)/4` = 90

For finding Q_{1} and Q_{3} 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 Q_{1} and Q_{2}.

∴ Q1 ≈ 1025 and Q3 ≈ 1248

∴ The limits of weight of middle 50% fishes lie between 1025 to 1248.