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Calculate Bowley’s coefficient of skewness Sk_{b} from the following data:

Marks above |
0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |

No. of students |
120 | 115 | 108 | 98 | 85 | 60 | 18 | 5 | 0 |

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#### Solution

To calculate Bowley’s coefficient of skewness Skb, we construct the following table:

Marks above |
No. of students ‘more than’ (c.f.) |
Marks |
Frequency (fi) |
Less than cumulative frequency (c.f.) |

0 | 120 | 0 – 10 | 5 | 5 |

10 | 115 | 10 – 20 | 7 | 12 |

20 | 108 | 20 – 30 | 10 | 22 |

30 | 98 | 30 – 40 | 13 | 35 ← Q_{1} |

40 | 85 | 40 – 50 | 25 | 60 ← Q_{2} |

50 | 60 | 50 – 60 | 42 | 102 ← Q_{3} |

60 | 18 | 60 – 70 | 13 | 115 |

70 | 5 | 70 – 80 | 5 | 120 |

80 | 0 | 80 – 90 | 0 | 120 |

Total |
120 |
– |

Here, N = 120

Q_{1 }class = class containing the `("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 30 – 40.

∴ L = 30, h = 10, f = 13, c.f. = 22

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

= `30 + 10/13(30 - 22)`

= `30 + 10/13(8)`

= 30 + 6.1538

∴ Q_{1} = 36.1538

Q_{2 }class = class containing the `("N"/2)^"th"` observation

∴ `"N"/2 = 120/2` = 60

Cumulative frequency which is just greater than (or equal to) 60 is 60.

∴ Q_{2} lies in the class 40 – 50.

∴ L = 40, h = 10, f = 25, c.f. = 35

∴ Q_{2} = `"L" + "h"/"f"("N"/2 - "c.f.")`

= `40 + 10/25 (60 - 35)`

= `40 + 10/25(25)`

∴ Q_{2 }= 50

Q_{3 }class = class containing the `((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 102.

∴ Q_{3} lies in the class 50 – 60.

∴ L = 50, h = 10, f = 42, c.f. = 60

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

= `50 + 10/42(90 - 60)`

= `50 + 10/42(30)`

= 50 + 7.1429

∴ Q_{3} = 57.1429

Bowley’s coefficient of skewness:

Sk_{b} = `("Q"_3 + "Q"_1 - 2"Q"_2)/("Q"_3 - "Q"_1)`

= `(57.1429 + 36.1538 - 2(50))/(57.1429 - 36.1538)`

= `(93.2967 - 100)/(20.9891)`

= `(-6.7033)/(20.9891)`

∴ Sk_{b} = – 0.3194

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