发布时间 : 星期六 文章(全国通用版)2019高考数学二轮复习 中档大题规范练(三)概率与统计 文更新完毕开始阅读
注:年份代码1~7分别对应年份2008~2014.
(1)由折线图看出,可用线性回归模型拟合y与t的关系,请用相关系数加以说明; (2)建立y关于t的回归方程(系数精确到0.01),预测2019年我国生活垃圾无害化处理量. 附注:
7
7
7
参考数据:?yi=9.32,?tiyi=40.17,i=1
i=1
? ?yi-yi=1
?=0.55,7≈2.646.
2
n? ?ti-t??yi-yi=1
?
,
参考公式:相关系数r=
n2
n? ?ti-t?? ?yi-yi=1
i=1
^
^
^
?
2
回归方程y =a +b t中斜率和截距的最小二乘估计公式分别为:
n? ?ti-t??yi-y^
?
^
^
i=1
b =
n,a =y-b t.
? ?ti-ti=1
?
2
解 (1)由折线图中数据和附注中参考数据得
7
7
2
t=4,? (ti-t)=28,
i=1
? ?yi-yi=1
7
?=0.55.
2
77
? (ti-t)(yi-y)=?tiyi-t?yi=40.17-4×9.32=2.89,
i=1
i=1
i=1
2.89所以r≈≈0.99.
0.55×2×2.646
因为y与t的相关系数近似为0.99,说明y与t的线性相关程度相当高,从而可以用线性回归模型拟合y与t的关系. 9.32
(2)由y=≈1.331及(1)得
7
1
7
? ?ti-t??yi-y?
^
i=1
b =
=2.89≈0.10, 7
28
? ?ti-t?
2
i=1
^
^
a =y-b t≈1.331-0.103×4≈0.92.
^
所以y关于t的线性回归方程为y =0.10t+0.92. 将2019年对应的t=12代入线性回归方程,得
^
y =0.92+0.10×12=2.12.
所以预测2019年我国生活垃圾无害化处理量将约为2.12亿吨.
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