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Calculate Karl Pearsonian’s coefficient of skewness Sk_{p} 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

The given table is the cumulative frequency table of more than type. From this table, we have to prepare the frequency distribution table and then calculate the value of Skp.

Construct the following table:

Mark above |
No. of students 'more than' (c.f.) |
Class-interval |
Frequencyf_{i} |
Mid valuex_{i} |
f_{i}x_{i} |
f_{i}x_{i}^{2} |

0 | 120 | 0 – 10 | 5 | 5 | 25 | 125 |

10 | 115 | 10 – 20 | 7 | 15 | 105 | 1575 |

20 | 108 | 20 – 30 | 10 | 25 | 250 | 6250 |

30 | 98 | 30 – 40 | 13 | 35 | 455 | 15925 |

40 | 85 | 40 – 50 | 25 | 45 | 1125 | 50625 |

50 | 60 | 50 – 60 | 42 | 55 | 2310 | 127050 |

60 | 18 | 60 – 70 | 13 | 65 | 845 | 54925 |

70 | 5 | 70 – 80 | 5 | 75 | 375 | 28125 |

80 | 0 | 80 – 90 | 0 | 85 | 0 | 0 |

Total |
120 |
– |
5490 |
284600 |

From the table, N = 120, `sumf_"i"x_"i" = 5490 "and" sumf_"i"x_"i"^2 = 284600`

Mean = `bar(x) - (sumf_"i"x_"i")/"N" = 5490/120` = 45.75

Maximum frequency 42 is of the class 50 – 60.

∴ Mode lies in the class 50 – 60.

∴ L = 50, f1 = 42, f0 = 25, f2 = 13, h = 10

∴ Mode = `"L" + ("f"_1 - "f"_0)/(2"f"_1 - "f"_0 - "f"_2) xx "h"`

= `50 + (42 - 25)/(2(42) - 25 - 13) xx 10`

= `50 + 17/(84 - 38) xx 10`

= `50 + 17/46 xx 10`

= 50 + 3.6957

= 53.6957

S.D. = `sqrt((sumf_"i"x_"i"^2)/"N" - (bar(x))^2`

= `sqrt(284600/120 - (45.75)^2`

= `sqrt(2371.6667 - 2093.0625)`

= `sqrt(278.6042)`

= 16.6914

Pearsonian’s coefficient of skewness:

Sk_{p} = `("Mean" - "Mode")/"S.D."`

= `(45.75 - 53.6957)/(16.6914)`

= `-(7.9457)/(16.6914)`

∴ Sk_{p} = – 0.4760**Alternate Method:**

Let u = `(x - 45)/10`

Mark above |
No. of students ‘more than’ (c.f.) |
Class |
Frequency(f_{i}) |
Mid valuex_{i} |
u_{i} |
f_{i}u_{i} |
f_{i}u_{i}^{2} |

0 | 120 | 0 –10 | 5 | 5 | − 4 | − 20 | 80 |

10 | 115 | 10 –20 | 7 | 15 | − 3 | − 21 | 63 |

20 | 108 | 20 – 30 | 10 | 25 | − 2 | − 20 | 40 |

30 | 98 | 30 –40 | 13 | 35 | − 1 | − 13 | 13 |

40 | 85 | 40 – 50 | 25 | 45 | 0 | 0 | 0 |

50 | 60 | 50 –60 | 42 | 55 | 1 | 42 | 42 |

60 | 18 | 60 –70 | 13 | 65 | 2 | 26 | 52 |

70 | 5 | 70 –80 | 5 | 75 | 3 | 15 | 45 |

80 | 0 | 80 –90 | 0 | 85 | 4 | 0 | 0 |

Total |
120 |
9 |
335 |

`bar(u) = (sumf_"i"u_"i")/"N" = 9/120` = 0.075

∴ `bar(x) = 45 + 10(bar(u))`

= 45 + 10(0.075)

= 45 + 0.75

= 45.75

Var(u) = `sigma_u^2 = (sumf_"i"u_"i"^2)/"N" - (bar(u))^2`

= `335/120 - (0.075)^2`

= 2.7917 – 0.0056

= 2.7861

Var(X) = h^{2} x Var(u) = 100 × 2.7861 = 278.61

S.D. = `sqrt(278.61)`

= 16.6916

Maximum frequency 42 is of the class 50 – 60.

∴ Mode lies in the class 50 – 60.

∴ L = 50, f1 = 42, f0 = 25, f2 = 13, h = 10

∴ Mode = `"L" + ("f"_1 - "f"_0)/(2"f"_1 - "f"_0 - "f"_2) xx "h"`

= `50 + (42 - 25)/(2(42) - 25 - 13) xx 10`

= `50 + 17/(84 - 38) xx 10`

= `50 + 17/46 xx 10`

= 50 + 3.6957

= 53.6957

∴ Sk_{p} = `("Mean"-"Mode")/"S.D."`

= `(45.75 - 53.6957)/(16.6916)`

= `(-7.9457)/(16.6916)`

= – 0.4760

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