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Calculate Q.D. of the following data:

Height of plants (in feet) |
2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 | 10 – 12 | 14 – 14 | 14 – 16 |

No. of plants |
15 | 20 | 25 | 12 | 18 | 13 | 17 |

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#### Solution

We construct the less than cumulative frequency table as follows:

Height of plants (in feet) |
No. of plants (f) |
Less than cumulative frequency (c.f.) |

2 – 4 | 15 | 15 |

4 – 6 | 20 | 35 ← Q_{1} |

6 – 8 | 25 | 60 |

8 – 10 | 12 | 72 |

10 – 12 | 18 | 90 ← Q_{3} |

12 – 14 | 13 | 103 |

14 – 16 | 17 | 120 |

Total |
N = 120 |

Here, N = 120

Q_{1} class = class containing `("N"/4)^"th"` observation

∴ `"N"/4 = 120/4` = 30

Cumulative frequency which is just greater than (or equal to) 30 is 35.

∴ Q_{1} lies in the class 4 – 6

∴ L = 4, c.f. = 15, f = 20, h = 2

∴ Q_{1} = `"L" + "h"/"f"("N"/4 - "c.f.")`

= `4 + 2/20 (30 - 15)`

= `4 + 1/10 xx 15`

= 4 + 1.5

= 5.5

Q_{3} class = class containing `((3"N")/4)^"th"` observation

∴ `(3"N")/4 = (3 xx 120)/4` = 90

Cumulative frequency which is just greater than (or equal to) 90 is 90.

∴ Q_{3} lies in the class 10 – 12

∴ L = 10, c.f. = 72, f = 18, h = 2

∴ Q_{3} = `"L" + "h"/"f"((3"N")/4 - "c.f.")`

= `10 + 2/18 (90 - 72)`

= `10 + 2/18 xx 18`

= 10 + 2

∴ Q_{3} = 12

∴ Q.D. = `("Q"_3 - "Q"_1)/2`

= `(12 - 5.5)/2`

= `(6.5)/2`

= 3.25