题目

Problem

The time taken, in seconds, to complete a puzzle was recorded for 160 pupils at a school. These times are summarised in the table.

time taken ( $t$ seconds)	$t\leq50$	$t\leq75$	$t\leq100$	$t\leq150$	$t\leq200$	$t\leq250$	$t\leq310$	$t\leq400$
cumulative frequency	8	30	62	102	128	143	154	160

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

(2)

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

(2)

题目中文翻译

某学校记录了 160 名学生完成一个谜题所用的时间（秒）。这些时间汇总如下表。

所用时间（ $t$ 秒）	$t\leq50$	$t\leq75$	$t\leq100$	$t\leq150$	$t\leq200$	$t\leq250$	$t\leq310$	$t\leq400$
累计频数	8	30	62	102	128	143	154	160

(a) 画出累计频数图表示这些数据。

(2)

(b) 利用你的图估计这些数据的四分位距。

(2)

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

展开

表中已经给出累计频数，直接在每个时间上界处描点，用平滑递增曲线连接，并标明时间单位和累计频数。

答题过程

展开

Plot the following cumulative frequency points:

(0,0),\ (50,8),\ (75,30),\ (100,62),\ (150,102),

(200,128),\ (250,143),\ (310,154),\ (400,160).

Join them with a smooth increasing curve. Label the axes “Time, $t$ (seconds)” and “Cumulative frequency”.

(b)

解法一

思路

展开

四分位数对应累计频数40及120。在图上从这两个高度横向读到曲线，再竖向读时间，最后相减。

答题过程

展开

There are 160 pupils, so read the lower quartile at cumulative frequency $\frac{160}{4}=40$ and the upper quartile at $\frac{3(160)}4=120$ .

Reading from the graph gives approximately

Q_1\approx80\text{ s},\qquad Q_3\approx185\text{ s}.

Thus the estimated interquartile range is

Q_3-Q_1\approx185-80=\boxed{105\text{ s}}.

CIE 9709 2026 June Paper 55 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程