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S1 2023 June Q1

A Level / Edexcel / S1

IAL 2023 June Paper · Question 1

题目

Problem

The histogram shows the distances, in km, that 274274 people travel to work.

Given that 6060 of these people travel between 1010 km and 2020 km to work, estimate

(a) the number of people who travel between 2222 km and 4545 km to work,

(3)

(b) the median distance travelled to work by these 274274 people,

(2)

(c) the mean distance travelled to work by these 274274 people.

(3)
题目中文翻译

直方图显示 274274 人上班路程的距离(km)。

已知其中 6060 人上班路程介于 1010 km 与 2020 km 之间,估计:

(a) 上班路程介于 2222 km 与 4545 km 之间的人数;

(b) 这 274274 人上班路程的中位数;

(c) 这 274274 人上班路程的平均数。

解答

(a)

解法一

思路

展开

先用 1010 到 2020 这一组的频数 6060 校准直方图比例。由图可读出各组频数为 100,70,60,20,24100,70,60,20,24,对应组距 00-55、55-1010、1010-2020、2020-3030、3030-6060。

答题过程

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From the histogram scale, the relevant frequencies are:

20⩽x<30: 20,30⩽x<60: 24.20\leqslant x<30:\ 20,\qquad 30\leqslant x<60:\ 24.

For 2222 to 3030,

810(20)=16.\begin{align*} \frac{8}{10}(20)=16. \end{align*}

For 3030 to 4545,

1530(24)=12.\begin{align*} \frac{15}{30}(24)=12. \end{align*}

Therefore the estimated number is

16+12=28.\begin{align*} 16+12=28. \end{align*}

(b)

解法一

思路

展开

中位数位置是第 137137 个。累计到 55 km 是 100100,累计到 1010 km 是 170170,所以中位数在 55 到 1010 这一组内。

答题过程

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The median is the 137137th value.

In the class 5⩽x<105\leqslant x<10, the frequency is 7070 and the class width is 55.

median= 5+137−10070×5= 7.642857…\begin{align*} \text{median} =&\,5+\frac{137-100}{70}\times5\\ =&\,7.642857\ldots \end{align*}

So the median distance is

7.64 km\begin{align*} 7.64\text{ km} \end{align*}

to 33 significant figures.

解法二

思路

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也可以用 n+12\frac{n+1}{2} 定位中位数。这里中位数位置为第 137.5137.5 个数据。

答题过程

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Using the n+1n+1 convention,

median= 5+137.5−10070×5= 7.67857…\begin{align*} \text{median} =&\,5+\frac{137.5-100}{70}\times5\\ =&\,7.67857\ldots \end{align*}

So the median distance is 7.68 km7.68\text{ km} to 33 significant figures.

(c)

解法一

思路

展开

用 grouped data 的平均数估计:每组用 midpoint 乘以 frequency,再除以总人数。

答题过程

展开

Using midpoints,

∑fx= 2.5(100)+7.5(70)+15(60) +25(20)+45(24)= 3255.\begin{align*} \sum fx =&\,2.5(100)+7.5(70)+15(60)\\ &\,\hspace{2pt}+25(20)+45(24)\\ =&\,3255. \end{align*}

Therefore

xˉ= 3255274= 11.879…\begin{align*} \bar{x} =&\,\frac{3255}{274}\\ =&\,11.879\ldots \end{align*}

The mean distance is

11.9 km\begin{align*} 11.9\text{ km} \end{align*}

to 33 significant figures.