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8. 某家全品类直播间希望根据用户的年消费额将其分成 2 个群体，分别是新粉和铁粉，以便于在购物节发送不同额度的消费券来维系客户.现技术人员随机提取 5 位客户的年消费额，如下表：
(1)将这 5 位客户分成两组，怎么分比较合理？请说明理由.
(2)随着直播间的爆火，客户人数的增多，主播希望分为新粉、铁粉、挚爱粉 3 组，以更好回馈粉丝，那么怎么分组比较合理？请说明理由.
(1)将这 5 位客户分成两组，怎么分比较合理？请说明理由.
(2)随着直播间的爆火，客户人数的增多，主播希望分为新粉、铁粉、挚爱粉 3 组，以更好回馈粉丝，那么怎么分组比较合理？请说明理由.
答案
8. (1)将数据排序得$2＜3＜4＜6＜7$,
|分类|$D^{2}_{1}$|5位客户消费额|$D^{2}_{2}$|$D^{2}_{1}+D^{2}_{2}$|
| ---- | ---- | ---- | ---- | ---- |
|第1种|$0$|$2$ $3$ $4$ $6$ $7$|$10$|$10$|
|第2种|$0.5$|$2$ $3$ $4$ $6$ $7$|$4.666667$|$5.166667$|
|第3种|$2$|$2$ $3$ $4$ $6$ $7$|$0.5$|$2.5$|
|第4种|$8.75$|$2$ $3$ $4$ $6$ $7$|$0$|$8.75$|
共有4种分组,由计算可以得到第3种分组方式组内离差平方和最小,比较合理.
(2)共有6种分组,根据组内离差平方和最小的原则,第2种和第4种分组都可以.理由如下:
|分类|$D^{2}_{1}$|5位客户消费额|$D^{2}_{2}$|$D^{2}_{3}$|$D^{2}_{1}+D^{2}_{2}+D^{2}_{3}$|
| ---- | ---- | ---- | ---- | ---- | ---- |
|第1种|$0$|$2$ $3$ $4$ $6$ $7$|$0$|$4.666667$|$4.666667$|
|第2种|$0$|$2$ $3$ $4$ $6$ $7$|$0.5$|$0.5$|$1$|
|第3种|$0$|$2$ $3$ $4$ $6$ $7$|$4.666667$|$0$|$4.666667$|
|第4种|$0.5$|$2$ $3$ $4$ $6$ $7$|$0$|$0.5$|$1$|
|第5种|$0.5$|$2$ $3$ $4$ $6$ $7$|$2$|$0$|$2.5$|
|第6种|$2$|$2$ $3$ $4$ $6$ $7$|$0$|$0$|$2$|
|分类|$D^{2}_{1}$|5位客户消费额|$D^{2}_{2}$|$D^{2}_{1}+D^{2}_{2}$|
| ---- | ---- | ---- | ---- | ---- |
|第1种|$0$|$2$ $3$ $4$ $6$ $7$|$10$|$10$|
|第2种|$0.5$|$2$ $3$ $4$ $6$ $7$|$4.666667$|$5.166667$|
|第3种|$2$|$2$ $3$ $4$ $6$ $7$|$0.5$|$2.5$|
|第4种|$8.75$|$2$ $3$ $4$ $6$ $7$|$0$|$8.75$|
共有4种分组,由计算可以得到第3种分组方式组内离差平方和最小,比较合理.
(2)共有6种分组,根据组内离差平方和最小的原则,第2种和第4种分组都可以.理由如下:
|分类|$D^{2}_{1}$|5位客户消费额|$D^{2}_{2}$|$D^{2}_{3}$|$D^{2}_{1}+D^{2}_{2}+D^{2}_{3}$|
| ---- | ---- | ---- | ---- | ---- | ---- |
|第1种|$0$|$2$ $3$ $4$ $6$ $7$|$0$|$4.666667$|$4.666667$|
|第2种|$0$|$2$ $3$ $4$ $6$ $7$|$0.5$|$0.5$|$1$|
|第3种|$0$|$2$ $3$ $4$ $6$ $7$|$4.666667$|$0$|$4.666667$|
|第4种|$0.5$|$2$ $3$ $4$ $6$ $7$|$0$|$0.5$|$1$|
|第5种|$0.5$|$2$ $3$ $4$ $6$ $7$|$2$|$0$|$2.5$|
|第6种|$2$|$2$ $3$ $4$ $6$ $7$|$0$|$0$|$2$|
9. 就数据个数 $ n = 5 $，$ k_{1} = 1 $，$ k_{2} = 2 $，$ k_{3} = 2 $，推导公式 $ D^{2}=D_{1}^{2}+D_{2}^{2}+D_{3}^{2}+[k_{1}(\overline{x}_{1}-\overline{x})^{2}+k_{2}(\overline{x}_{2}-\overline{x})^{2}+k_{3}(\overline{x}_{3}-\overline{x})^{2}] $.
答案
9. $D^{2}=(x_{1}-\overline{x})^{2}+(x_{2}-\overline{x})^{2}+(x_{3}-\overline{x})^{2}+(x_{4}-\overline{x})^{2}+(x_{5}-\overline{x})^{2}$
$=(x_{1}-\overline{x}_{1}+\overline{x}_{1}-\overline{x})^{2}+(x_{2}-\overline{x}_{2}+\overline{x}_{2}-\overline{x})^{2}+(x_{3}-\overline{x}_{2}+\overline{x}_{2}-\overline{x})^{2}+(x_{4}-\overline{x}_{3}+\overline{x}_{3}-\overline{x})^{2}+(x_{5}-\overline{x}_{3}+\overline{x}_{3}-\overline{x})^{2}$
$=\left \lbrack(x_{1}-\overline{x}_{1})^{2}+2(x_{1}-\overline{x}_{1})(\overline{x}_{1}-\overline{x})+(\overline{x}_{1}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{2}-\overline{x}_{2})^{2}+2(x_{2}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+(\overline{x}_{2}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{3}-\overline{x}_{2})^{2}+2(x_{3}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+(\overline{x}_{2}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{4}-\overline{x}_{3})^{2}+2(x_{4}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})+(\overline{x}_{3}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{5}-\overline{x}_{3})^{2}+2(x_{5}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})+(\overline{x}_{3}-\overline{x})^{2}\right \rbrack$
$=\left \lbrack(x_{1}-\overline{x}_{1})^{2}+(x_{2}-\overline{x}_{2})^{2}+(x_{3}-\overline{x}_{2})^{2}+(x_{4}-\overline{x}_{3})^{2}+(x_{5}-\overline{x}_{3})^{2}\right \rbrack+\left \lbrack2(x_{1}-\overline{x}_{1})(\overline{x}_{1}-\overline{x})+2(x_{2}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+2(x_{3}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+2(x_{4}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})+2(x_{5}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})\right \rbrack+\left \lbrack(\overline{x}_{1}-\overline{x})^{2}+(\overline{x}_{2}-\overline{x})^{2}+(\overline{x}_{2}-\overline{x})^{2}+(\overline{x}_{3}-\overline{x})^{2}+(\overline{x}_{3}-\overline{x})^{2}\right \rbrack$
$=(x_{1}-\overline{x}_{1})^{2}+\left \lbrack(x_{2}-\overline{x}_{2})^{2}+(x_{3}-\overline{x}_{2})^{2}\right \rbrack+\left \lbrack(x_{4}-\overline{x}_{3})^{2}+(x_{5}-\overline{x}_{3})^{2}\right \rbrack+2(x_{1}-\overline{x}_{1})(\overline{x}_{1}-\overline{x})+2(\overline{x}_{2}-\overline{x})(x_{2}+x_{3}-2\overline{x}_{2})+2(\overline{x}_{3}-\overline{x})(x_{4}+x_{5}-2\overline{x}_{3})]+(\overline{x}_{1}-\overline{x})^{2}+2(\overline{x}_{2}-\overline{x})^{2}+2(\overline{x}_{3}-\overline{x})^{2}$
$=(D^{2}_{1}+D^{2}_{2}+D^{2}_{3})+0+\left \lbrack k_{1}(\overline{x}_{1}-\overline{x})^{2}+k_{2}(\overline{x}_{2}-\overline{x})^{2}+k_{3}(\overline{x}_{3}-\overline{x})^{2}\right \rbrack$.
$=(x_{1}-\overline{x}_{1}+\overline{x}_{1}-\overline{x})^{2}+(x_{2}-\overline{x}_{2}+\overline{x}_{2}-\overline{x})^{2}+(x_{3}-\overline{x}_{2}+\overline{x}_{2}-\overline{x})^{2}+(x_{4}-\overline{x}_{3}+\overline{x}_{3}-\overline{x})^{2}+(x_{5}-\overline{x}_{3}+\overline{x}_{3}-\overline{x})^{2}$
$=\left \lbrack(x_{1}-\overline{x}_{1})^{2}+2(x_{1}-\overline{x}_{1})(\overline{x}_{1}-\overline{x})+(\overline{x}_{1}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{2}-\overline{x}_{2})^{2}+2(x_{2}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+(\overline{x}_{2}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{3}-\overline{x}_{2})^{2}+2(x_{3}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+(\overline{x}_{2}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{4}-\overline{x}_{3})^{2}+2(x_{4}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})+(\overline{x}_{3}-\overline{x})^{2}\right \rbrack+\left \lbrack(x_{5}-\overline{x}_{3})^{2}+2(x_{5}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})+(\overline{x}_{3}-\overline{x})^{2}\right \rbrack$
$=\left \lbrack(x_{1}-\overline{x}_{1})^{2}+(x_{2}-\overline{x}_{2})^{2}+(x_{3}-\overline{x}_{2})^{2}+(x_{4}-\overline{x}_{3})^{2}+(x_{5}-\overline{x}_{3})^{2}\right \rbrack+\left \lbrack2(x_{1}-\overline{x}_{1})(\overline{x}_{1}-\overline{x})+2(x_{2}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+2(x_{3}-\overline{x}_{2})(\overline{x}_{2}-\overline{x})+2(x_{4}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})+2(x_{5}-\overline{x}_{3})(\overline{x}_{3}-\overline{x})\right \rbrack+\left \lbrack(\overline{x}_{1}-\overline{x})^{2}+(\overline{x}_{2}-\overline{x})^{2}+(\overline{x}_{2}-\overline{x})^{2}+(\overline{x}_{3}-\overline{x})^{2}+(\overline{x}_{3}-\overline{x})^{2}\right \rbrack$
$=(x_{1}-\overline{x}_{1})^{2}+\left \lbrack(x_{2}-\overline{x}_{2})^{2}+(x_{3}-\overline{x}_{2})^{2}\right \rbrack+\left \lbrack(x_{4}-\overline{x}_{3})^{2}+(x_{5}-\overline{x}_{3})^{2}\right \rbrack+2(x_{1}-\overline{x}_{1})(\overline{x}_{1}-\overline{x})+2(\overline{x}_{2}-\overline{x})(x_{2}+x_{3}-2\overline{x}_{2})+2(\overline{x}_{3}-\overline{x})(x_{4}+x_{5}-2\overline{x}_{3})]+(\overline{x}_{1}-\overline{x})^{2}+2(\overline{x}_{2}-\overline{x})^{2}+2(\overline{x}_{3}-\overline{x})^{2}$
$=(D^{2}_{1}+D^{2}_{2}+D^{2}_{3})+0+\left \lbrack k_{1}(\overline{x}_{1}-\overline{x})^{2}+k_{2}(\overline{x}_{2}-\overline{x})^{2}+k_{3}(\overline{x}_{3}-\overline{x})^{2}\right \rbrack$.
