【答案】
分析:對(duì)新數(shù)據(jù)按大小排列,然后根據(jù)平均數(shù)和中位數(shù)的定義計(jì)算即可.
解答:解:由平均數(shù)定義可知:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162038629269438/SYS201310221620386292694003_DA/0.png)
(a
1+a
2+a
3+0+a
4+a
5)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162038629269438/SYS201310221620386292694003_DA/1.png)
×5a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162038629269438/SYS201310221620386292694003_DA/2.png)
a;
將這組數(shù)據(jù)按從小到大排列為0,a
5,a
4,a
3,a
2,a
1;由于有偶數(shù)個(gè)數(shù),取最中間兩個(gè)數(shù)的平均數(shù).
∴其中位數(shù)為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022162038629269438/SYS201310221620386292694003_DA/3.png)
.
故選D.
點(diǎn)評(píng):本題考查了平均數(shù)和中位數(shù)的定義.平均數(shù)是指在一組數(shù)據(jù)中所有數(shù)據(jù)之和再除以數(shù)據(jù)的個(gè)數(shù);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(gè)數(shù)有關(guān),因此求一組數(shù)據(jù)的中位數(shù)時(shí),先將該組數(shù)據(jù)按從小到大(或按從大到小)的順序排列,然后根據(jù)數(shù)據(jù)的個(gè)數(shù)確定中位數(shù):當(dāng)數(shù)據(jù)個(gè)數(shù)為奇數(shù)時(shí),則中間的一個(gè)數(shù)即為這組數(shù)據(jù)的中位數(shù);當(dāng)數(shù)據(jù)個(gè)數(shù)為偶數(shù)時(shí),則最中間的兩個(gè)數(shù)的算術(shù)平均數(shù)即為這組數(shù)據(jù)的中位數(shù).