题目

Problem

The times of 240 competitors taking part in an event are recorded correct to the nearest minute. The results are summarised in the table.

Time (minutes)	1–10	11–20	21–25	26–30	31–50
Frequency	12	38	68	76	46

(a) Draw a histogram to represent this information.

(4)

(b) Calculate an estimate of the mean time taken by the 240 competitors.

(3)

题目中文翻译

记录了参加某项比赛的 240 名选手所用的时间，并取至最接近的分钟数。结果汇总如下表。

时间（分钟）	1–10	11–20	21–25	26–30	31–50
频数	12	38	68	76	46

(a) 画出直方图表示这些数据。

(4)

(b) 计算这 240 名选手所用时间的平均数的估计值。

(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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记录到最接近的分钟，组界需要加减0.5。组距不等，因此用频率密度作为柱高。

答题过程

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Use continuous class boundaries and calculate frequency density as frequency divided by class width.

Class boundaries (minutes)	Width	Frequency density
$0.5$ to $10.5$	$10$	$12/10=1.2$
$10.5$ to $20.5$	$10$	$38/10=3.8$
$20.5$ to $25.5$	$5$	$68/5=13.6$
$25.5$ to $30.5$	$5$	$76/5=15.2$
$30.5$ to $50.5$	$20$	$46/20=2.3$

Draw adjoining bars with these heights. Label the axes “Time (minutes)” and “Frequency density”.

(b)

解法一

思路

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每组取中点代表该组数据，以频数加权求平均。保留准确中间值，最后按精度要求给答案。

答题过程

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The class midpoints are $5.5,15.5,23,28,40.5$ minutes. Hence

\begin{align*} \sum fm=&\,12(5.5)+38(15.5)+68(23)\\[4mm] &\,\hspace{2pt}+76(28)+46(40.5)\\[4mm] =&\,6210. \end{align*}

The estimated mean time is

\frac{6210}{240}=25.875 \approx\boxed{25.9\text{ minutes}}\quad\text{(3 s.f.)}.

CIE 9709 2025 November Paper 52 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程