【答案】
分析:根據(jù)平均數(shù)公式與方差公式即可求解.
解答:解:∵據(jù)x
1,x
2,x
3,x
4,x
5的平均數(shù)是2,
∴
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=2,
∵數(shù)據(jù)x
1,x
2,x
3,x
4,x
5的平均數(shù)是2,方差是
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,
∴
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[(x
1-2)
2+(x
2-2)
2+[(x
3-2)
2+(x
4-2)
2+(x
5-2)
2]=
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①;
∴3x
1-2,3x
2-2,3x
3-2,3x
4-2,3x
5-2,的平均數(shù)是
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,
=3×
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-2=4.
∴
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[(3x
1-2-4)
2+(3x
2-2-4)
2+(3x
3-2-4)
2+(3x
4-2-4)
2+(3x
5-2-4)
2]
=
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[9(x
1-2)
2+9(x
2-2)
2+9(x
3-2)
2+9(x
4-2)
2+9(x
5-2)
2]
=
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×9[(x
1-2)
2+(x
2-2)
2+(x
3-2)
2+(x
4-2)
2+(x
5-2)
2]②
把①代入②得,方差是:
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×9=3.
故答案為:4;3.
點(diǎn)評(píng):本題考查了平均數(shù)的計(jì)算公式和方差的定義:一般地設(shè)n個(gè)數(shù)據(jù),x
1,x
2,…x
n的平均數(shù)為
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,則S
2=
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[(x
1-
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)
2+(x
2-
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)
2+…+(x
n-
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)
2],它反映了一組數(shù)據(jù)的波動(dòng)大小,方差越大,波動(dòng)性越大,反之也成立.