Calculate Q_{1}, D_{6,} and P_{15} for the following data:

Mid value |
25 | 75 | 125 | 175 | 225 | 275 |

Frequency |
10 | 70 | 80 | 100 | 150 | 90 |

#### Solution

Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.

∴ the class intervals will be 0 – 50, 50 – 100, etc.

We construct the less than cumulative frequency table as given below:

Class interval |
Frequency (f) |
Less than cumulative frequency (c.f.) |

0 – 50 | 10 | 10 |

50 – 100 | 70 | 80 ← P_{15} |

100 – 150 | 80 | 160 ← Q_{1} |

150 – 200 | 100 | 260 |

200 – 250 | 150 | 410 ← D_{6} |

250 – 300 | 90 | 500 |

Total |
500 |

Here, N = 500

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

∴ `"N"/4 = 500/4` = 125

Cumulative frequency which is just greater than (or equal) to 125 is 160.

∴ Q_{1} lies in the class 100 – 150.

∴ L = 100, h = 50, f = 80, c.f. = 80

∴ Q_{1} = `"L"+"h"/"f"("N"/4-"c.f.")`

= `100 + 50/80 (125 - 80)`

= `100 + 5/8 (45)`

= 100 + 28.125

= 128.125

D_{6} class = class containing `((6"N")/10)^"th"` observation

∴ `(6"N")/10=(6xx500)/10` = 300

Cumulative frequency which is just greater than (or equal) to 300 is 410.

∴ D_{6} lies in the class 200 – 250

∴ L = 200, h = 50, f = 150, c.f. = 260

∴ D_{6} = `"L"+"h"/"f"((6"N")/4-"c.f.")`

= `200 + 50/150 (300 - 260)`

= `200 + 1/3 (40)`

= 200 + 13.33

= 213.33

∴ P_{15} class = class containing `((15"N")/100)^"th"` observation

∴ `(15"N")/100 =(15 xx 500)/100` = 75

Cumulative frequency which is just greater than (or equal) to 75 is 80.

∴ P_{15} lies in the class 50 – 100

∴ L = 50, h = 50, f = 70, c.f. = 10

∴ P_{15} = `"L"+"h"/"f"((15"N")/100-"c.f.")`

= `50 + 50/70 (75 - 10)`

= `50 + 5/7 (65)`

= `50 + 325/7`

= 50 + 46.4286

= 96.4286

∴ Q_{1} = 128.125, D_{6} = 213.33, P_{15} = 96.4286