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S1 2019 Oct Q4

A Level / Edexcel / S1

IAL 2019 Oct Paper · Question 4

题目

Problem

A random sample of 10 boys A, B, C, D, E, F, G, H, I and J is taken from a junior athletics club. Each boy selected is asked to run a 100 metre race and a 200 metre race. The time taken, xx seconds, by each boy to run the 100 metre race is recorded and the time taken, yy seconds, by each boy to run the 200 metre race is recorded. The results are plotted on the scatter diagram below.

(a) State, without calculation, which of the 3 values below is most likely to be a value of the product moment correlation coefficient for the data in the scatter diagram.

0.720.050.950.72\qquad 0.05\qquad 0.95
(1)

In the sample of 10 boys, one is a junior champion 100 metre runner and one is a junior champion 200 metre runner.

(b) Write down the boy who is most likely to be the 100 metre junior champion.

(1)

The data for the two junior champions are removed and the remaining data are summarised below

∑x=164.4∑x2=3445.26Syy=67.52Sxy=60.85\sum x=164.4\qquad \sum x^2=3445.26\qquad S_{yy}=67.52\qquad S_{xy}=60.85

(c) (i) Calculate the value of the product moment correlation coefficient for the remaining data.

(3)

(ii) Comment, in context, on the value of the product moment correlation coefficient that you obtained in part (i).

(1)
题目中文翻译

从一个青少年田径俱乐部随机抽取 10 名男生 A、B、C、D、E、F、G、H、I、J。每人分别参加 100 米和 200 米赛跑,100 米用时记为 xx 秒,200 米用时记为 yy 秒,结果绘于散点图中。

(a) 不作计算,从所给三个数值中选出最可能是该散点图数据的积矩相关系数。

样本中一人是青少年 100 米冠军,另一人是青少年 200 米冠军。

(b) 写出最可能是青少年 100 米冠军的男生。

移除两位青少年冠军的数据后,余下数据的汇总量如题。

(c) (i) 计算余下数据的积矩相关系数;(ii) 结合情境评论该相关系数。

解答

(a)

解法一

思路

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图中整体是正相关,但有两个明显偏离主趋势的点,所以相关性不会接近 0.95,最合理是 0.72。

答题过程

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The most likely value is

0.72.\begin{align*} 0.72. \end{align*}

(b)

解法一

思路

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100 metre champion 应该有最小的 100 metre time,也就是图中 xx 值最小的点。

答题过程

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The boy most likely to be the 100 metre junior champion is

C.C.

(c)

(i)

思路

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剩下 8 个数据。先用 ∑x\sum x 和 ∑x2\sum x^2 求 SxxS_{xx},再代入 rr 的公式。

答题过程

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For the remaining 8 boys,

Sxx=∑x2−(∑x)2n.\begin{align*} S_{xx}=\sum x^2-\frac{(\sum x)^2}{n}. \end{align*}

So

Sxx=3445.26−164.428=66.84.\begin{align*} S_{xx}=3445.26-\frac{164.4^2}{8}=66.84. \end{align*}

Therefore

r= SxySxxSyy= 60.8566.84(67.52)= 0.90578….\begin{align*} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\ =&\,\frac{60.85}{\sqrt{66.84(67.52)}}\\ =&\,0.90578\ldots. \end{align*}

Hence

r=0.906.\begin{align*} r=0.906. \end{align*}

(ii)

思路

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正的且接近 1 的相关系数表示两项赛跑用时有强正线性相关;应结合“用时越短即越快”来解释。

答题过程

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This shows that boys who are faster in the 100 metres also tend to be faster in the 200 metres.