📜  统计-四分位数偏差

📅  最后修改于: 2021-01-23 06:49:40             🧑  作者: Mango


它取决于较低的四分位数$ {Q_1} $和较高的四分位数$ {Q_3} $。差异$ {Q_3-Q_1} $被称为四分位间距。差$ {Q_3-Q_1} $除以2称为半中间四分位数范围或四分位数偏差。

$ {QD = \ frac {Q_3-Q_1} {2}} $

四分位数偏差系数

基于四分位数偏差的色散的相对度量称为四分位数偏差系数。它的特征是

$ {系数\ of \四分位数\偏差\ = \ frac {Q_3-Q_1} {Q_3 + Q_1}} $

问题陈述:

根据以下数据计算四分位数偏差和四分位数偏差系数:

Maximum Load
(short-tons)
Number of Cables
9.3-9.7 22
9.8-10.2 55
10.3-10.7 12
10.8-11.2 17
11.3-11.7 14
11.8-12.2 66
12.3-12.7 33
12.8-13.2 11

解:

Maximum Load
(short-tons)
Number of Cables
(f)
Class
Bounderies
Cumulative
Frequencies
9.3-9.7 2 9.25-9.75 2
9.8-10.2 5 9.75-10.25 2 + 5 = 7
10.3-10.7 12 10.25-10.75 7 + 12 = 19
10.8-11.2 17 10.75-11.25 19 + 17 = 36
11.3-11.7 14 11.25-11.75 36 + 14 = 50
11.8-12.2 6 11.75-12.25 50 + 6 = 56
12.3-12.7 3 12.25-12.75 56 + 3 = 59
12.8-13.2 1 12.75-13.25 59 + 1 = 60

$ {Q_1} $

$ {\ frac {n} {4} ^ {th}} $项的价值= $ {\ frac {60} {4} ^ {th}} $项的价值= $ {15 ^ {th}} $项。因此$ {Q_1} $位于10.25-10.75类中。

$ {Q_1 = 1+ \ frac {h} {f}(\ frac {n} {4}-c)\\ [7pt] \,其中\ l = 10.25,\ h = 0.5,\ f = 12,\ \ frac {n} {4} = 15 \和\ c = 7,\\ [7pt] \,= 10.25+ \ frac {0.5} {12}(15-7),\\ [7pt] \,= 10.25 +0.33,\\ [7pt] \,= 10.58} $

$ {Q_3} $

$ {\ frac {3n} {4} ^ {th}} $项目的价值= $ {\ frac {3 \ times 60} {4} ^ {th}} $的价值= $ {45 ^ {th} } $个项目。因此,$ {Q_3} $位于类11.25-11.75中。

$ {Q_3 = 1+ \ frac {h} {f}(\ frac {3n} {4}-c)\\ [7pt] \,其中\ l = 11.25,\ h = 0.5,\ f = 14,\ \ frac {3n} {4} = 45 \和\ c = 36,\\ [7pt] \,= 11.25+ \ frac {0.5} {14}(45-36),\\ [7pt] \,= 11.25 +0.32,\\ [7pt] \,= 11.57} $

四分位偏差

$ {QD = \ frac {Q_3-Q_1} {2} \\ [7pt] \,= \ frac {11.57-10.58} {2},\\ [7pt] \,= \ frac {0.99} {2}, \\ [7pt] \,= 0.495} $

四分位数偏差系数

$ {系数\ of \四分位数\偏差\ = \ frac {Q_3-Q_1} {Q_3 + Q_1} \\ [7pt] \,= \ frac {11.57-10.58} {11.57 + 10.58},\\ [7pt] \ ,= \ frac {0.99} {22.15},\\ [7pt] \,= 0.045} $