有一組數(shù)據(jù)11,8,10,9,12的極差是 ,方差是 .
【答案】
分析:極差是數(shù)據(jù)中最大數(shù)與最小數(shù)的差,此數(shù)據(jù)中最大數(shù)是12,最小數(shù)是8,所以極差是把兩數(shù)相減即可;要求方差,首先求這組數(shù)據(jù)的平均數(shù),求出平均數(shù)后,再利用方差公式方差公式S
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/0.png)
[(x
1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/1.png)
)
2+[(x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/2.png)
)
2+…+[(x
n-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/3.png)
)
2],代入數(shù)據(jù)求出即可.
解答:解;極差是;12-8=4;
平均數(shù):
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/4.png)
=(11+8+10+9+12)÷5=10
方差:
S
2=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/5.png)
[(x
1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/6.png)
)
2+[(x
2-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/7.png)
)
2+…+[(x
n-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/8.png)
)
2],
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/9.png)
[(11-10)
2+(8-10)
2+(10-10)
2+(9-10)
2+(12-10)
2]
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103200058014102338/SYS201311032000580141023012_DA/10.png)
(1+4+0+1+4),
=2,
故答案為:4,2.
點評:此題主要考查了極差與方差的有關知識,方差大小代表數(shù)據(jù)的波動大小,方差越大代表這組數(shù)據(jù)波動越大,方差越小波動越小,極差則是最值之間的差值,方差與極差在中考中是熱點問題.