Find graphically the values of D_{3} and P_{65} for the data given below:

I.Q of students |
60 – 69 | 70 – 79 | 80 – 89 | 90 – 99 | 100 – 109 | 110 – 119 | 120 – 129 |

No. of students |
20 | 40 | 50 | 50 | 20 | 10 | 10 |

#### Solution

Since the given data is not continuous, we have to convert it in the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval. To draw a ogive curve, we construct the less than cumulative frequency table as given below:

I.Q. of students |
No. of students(f) |
Less than cumulative frequency(c.f.) |

59.5 – 69.5 | 20 | 20 |

69.5 – 79.5 | 40 | 60 |

79.5 – 89.5 | 50 | 110 |

89.5 – 99.5 | 50 | 160 |

99.5 – 109.5 | 20 | 180 |

109.5 – 119.5 | 10 | 190 |

119.5 – 129.5 | 10 | 200 |

Total |
200 |

Points to be plotted are (69.5, 20), (79.5, 60), (89.5, 110), (99.5, 160), (109.5, 180), (119.5, 190), (129.5, 200).

N = 200

For D_{3}, `(3"N")/10=(3xx200)/10` = 60

For P_{65}, `(65"N")/100=(65xx200)/100` = 130

∴ We take the values 60 and 130 on the Y-axis. From these points we draw lines parallel to X-axis and from the points where these lines intersect less than ogive, we draw perpendiculars on X-axis. The foot of perpendiculars represents the median of the values, D_{3,} and P_{65}.

∴ D_{3} = 79.5, P_{65} = 93.5