已知一組數(shù)據(jù)1,2,1,0,-1,-2,0,-1,則這組數(shù)據(jù)的平均數(shù)為 ,中位數(shù)為 ,方差為 .
【答案】
分析:根據(jù)平均數(shù),中位數(shù)定義及方差公式求解.
方差公式:S
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/0.png)
[(x
1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/1.png)
)
2+(x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/2.png)
)
2+…+(x
n-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/3.png)
)
2].
解答:解:平均數(shù)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/4.png)
(1+2+1+0-1-2+0-1)=0,
排序后第4和第5個數(shù)的平均數(shù)為0,即中位數(shù)為0,
方差為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/6.png)
.
故填0,0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/7.png)
.
點評:本題考查平均數(shù)、中位數(shù)和方差的概念.
一組數(shù)據(jù)的中位數(shù)與這組數(shù)據(jù)的排序及數(shù)據(jù)個數(shù)有關(guān),因此求一組數(shù)據(jù)的中位數(shù)時,先將該組數(shù)據(jù)按從小到大(或按從大到�。┑捻樞蚺帕校缓蟾鶕�(jù)數(shù)據(jù)的個數(shù)確定中位數(shù):當數(shù)據(jù)個數(shù)為奇數(shù)時,則中間的一個數(shù)即為這組數(shù)據(jù)的中位數(shù);當數(shù)據(jù)個數(shù)為偶數(shù)時,則最中間的兩個數(shù)的算術(shù)平均數(shù)即為這組數(shù)據(jù)的中位數(shù).
方差公式為:S
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/8.png)
[(x
1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/9.png)
)
2+(x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/10.png)
)
2+…+(x
n-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101191146902988947/SYS201311011911469029889003_DA/11.png)
)
2].