• 3. 计算公式为( 1) ( 2) ( )2n n n nx x xxn nα α ααα+ + −+ + +− ×=\uf04c( 1) ( 2) ( )2n n n nx x xxn nα α ααα+ + −+ + +− ×=\uf04c102α ≤ ≤102α ≤ ≤n 表示观察值的个数; α α 表示切尾系数, 切尾均值(例题分析)\uf06f 【例】某次比赛共有11名评委,对某位歌手的给分分别是:经整理得到顺序统计量值为1 2 3 4 5 6, , , , , ,9.22,9.25,9.20,9.30,9.65,9.30,x x x x x x7 8 9 10 11, , , , 9.27,9.20,9.28,9.25,9.24x x x x x(1) (2) (3) (4) (5) (6), , , , , ,9.20, 9.20, 9.22, 9.24, 9.25, 9.25,x x x x x x(7) (8) (9) (10) (11), , , , 9.27, 9.28, 9.30, 9.30, 9.65x x x x x去掉一个最高分和一个最低分,取 α = 1/11( ) ( ) ( )11 1/11 1 11 1/11 2 11 11 1/111/11(2) (3) (10)11 2 11 1/119.2 9.22 9.3 9.2611 2 9x x xxx x x× + × + − ×+ + +=− × ×+ + ++ + += = =−\uf04c\uf04c\uf04c( ) ( ) ( )11 1/11 1 11 1/11 2 11 11 1/111/11(2) (3) (10)11 2 11 1/119.2 9.22 9.3 9.2611 2 9x x xxx x x× + × + − ×+ + +=− × ×+ + ++ + += = =−\uf04c\uf04c\uf04c 切尾均值的特例如果极端分数不止一个,且数量较大,可以去掉两个最高分和两个最低分,此时切尾系数为2/11 ,依次类推。,依次类推。(1) (2) (11)0/11(6)5/1160,115/11, 1/2)1x x xxxx xαα α+ + += == = = →\uf04c即算术平均数即中位数((1) (2) (11)0/11(6)5/1160,115/11, 1/2)1x x xxxx xαα α+ + += == = = →\uf04c