题目

Problem

The time taken, in minutes, to walk to school was recorded for 200 pupils at a certain school. These times are summarised in the following table.

Time taken ( $t$ minutes)	$t\leq 15$	$t\leq 25$	$t\leq 30$	$t\leq 40$	$t\leq 50$	$t\leq 70$
Cumulative frequency	18	46	88	140	176	200

(a) Draw a cumulative frequency graph to illustrate the data.

(2)

(b) Use your graph to estimate the median and the interquartile range of the data.

(3)

题目中文翻译

记录某所学校 200 名学生步行上学所用的时间，单位为分钟。这些用时汇总如下表。

用时（ $t$ 分钟）	$t\leq 15$	$t\leq 25$	$t\leq 30$	$t\leq 40$	$t\leq 50$	$t\leq 70$
累积频数	18	46	88	140	176	200

(a) 画出累积频数图来表示这些数据。

(2)

(b) 利用你所画的图，估计这些数据的中位数和四分位距。

(3)

原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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表内已经是累计频数，用组上界描点，并从原点画递增曲线。

答题过程

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Plot the points

(0,0),\ (15,18),\ (25,46),\ (30,88),

(40,140),\ (50,176),\ (70,200).

Use linear scales with time $t$ (minutes) on the horizontal axis and cumulative frequency on the vertical axis. Join the points with a smooth increasing curve.

(b)

解法一

思路

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200人的中位数、下四分位数、上四分位数分别对应累计频数100、50、150。

答题过程

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Read horizontally from cumulative frequencies 50, 100 and 150 to the curve, then vertically down to the time axis. The estimated quartiles and median are

Q_1\approx26,\qquad \text{median}\approx33,\qquad Q_3\approx42.

Thus the median is approximately 33 minutes, and

\mathrm{IQR}\approx42-26=\boxed{16\text{ minutes}}.

(c)

解法一

思路

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累计频数作差得到各组人数，再用组中点估计每组的用时。

答题过程

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The class frequencies and midpoints are

Time interval (minutes)	Frequency $f$	Midpoint $m$
$0<t\leq15$	$18$	$7.5$
$15<t\leq25$	$28$	$20$
$25<t\leq30$	$42$	$27.5$
$30<t\leq40$	$52$	$35$
$40<t\leq50$	$36$	$45$
$50<t\leq70$	$24$	$60$

Hence

\begin{align*} \sum fm=&\,18(7.5)+28(20)+42(27.5)\\[4mm] &\,\hspace{2pt}+52(35)+36(45)+24(60)\\[4mm] =&\,6730. \end{align*}

The estimated mean time is

\frac{6730}{200}=33.65\approx\boxed{33.7\text{ minutes}}.

CIE 9709 2024 November Paper 51 Q3

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