题目

Problem

The histogram shows the distances, in km, that $274$ people travel to work.

Histogram

Given that $60$ of these people travel between $10$ km and $20$ km to work, estimate

(a) the number of people who travel between $22$ km and $45$ km to work,

(3)

(b) the median distance travelled to work by these $274$ people,

(2)

(3)

解答

(a)

解法一

思路

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先用 $10$ 到 $20$ 这一组的频数 $60$ 校准直方图比例。由图可读出各组频数为 $100,70,60,20,24$ ，对应组距 $0$ - $5$ 、 $5$ - $10$ 、 $10$ - $20$ 、 $20$ - $30$ 、 $30$ - $60$ 。

答题过程

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From the histogram scale, the relevant frequencies are:

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

For $22$ to $30$ ,

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

For $30$ to $45$ ,

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

Therefore the estimated number is

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

(b)

解法一

思路

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中位数位置是第 $137$ 个。累计到 $5$ km 是 $100$ ，累计到 $10$ km 是 $170$ ，所以中位数在 $5$ 到 $10$ 这一组内。

答题过程

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The median is the $137$ th value.

In the class $5\leqslant x<10$ , the frequency is $70$ and the class width is $5$ .

\begin{align*} \text{median} =&\,5+\frac{137-100}{70}\times5\\[3mm] =&\,7.642857\ldots \end{align*}

So the median distance is

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

to $3$ significant figures.

(c)

解法一

思路

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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

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Using midpoints,

\begin{align*} \sum fx =&\,2.5(100)+7.5(70)+15(60)\\[3mm] &\,\hspace{2pt}+25(20)+45(24)\\[3mm] =&\,3255. \end{align*}

Therefore

\begin{align*} \bar{x} =&\,\frac{3255}{274}\\[3mm] =&\,11.879\ldots \end{align*}

The mean distance is

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

to $3$ significant figures.

IAL 2023 June Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程