题目

Problem

On a particular day, the times taken to travel to Meriday college by 150 students are recorded, correct to the nearest minute. These times are summarised in the table.

time, in minutes	5–9	10–14	15–19	20–29	30–39	40–49	50–59	60–79
frequency	4	11	14	26	48	28	15	4

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

(4)

(b) Use your graph to estimate the 60th percentile of the times taken to travel to college by these 150 students.

(2)

(c) Calculate an estimate for the mean of the times taken to travel to college by these 150 students.

(3)

题目中文翻译

某一天，记录了 150 名学生前往 Meriday 学院所用的时间，并取至最接近的分钟数。这些时间汇总如下表。

时间（分钟）	5–9	10–14	15–19	20–29	30–39	40–49	50–59	60–79
频数	4	11	14	26	48	28	15	4

(a) 画出累计频数图表示这些数据。

(4)

(b) 利用你的图估计这 150 名学生前往学院所用时间的第 60 百分位数。

(2)

(c) 计算这 150 名学生前往学院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

展开

时间记录到最接近的分钟，因此画累计频数时应使用连续组界，例如9.5而不是9。先累加频数，再在各组上界描点。

答题过程

展开

The cumulative frequencies are

4,\ 15,\ 29,\ 55,\ 103,\ 131,\ 146,\ 150.

Since times are recorded to the nearest minute, plot at the upper class boundaries:

(4.5,0),\ (9.5,4),\ (14.5,15),\ (19.5,29),

(29.5,55),\ (39.5,103),\ (49.5,131),

(59.5,146),\ (79.5,150).

Join the points with a smooth increasing curve. Label the axes “Time (minutes)” and “Cumulative frequency”.

(b)

解法一

思路

展开

第60百分位对应累计频数90。在图上从90横向移到曲线，再向下读出时间。

答题过程

展开

The required cumulative frequency is

0.60\times150=90.

Draw a horizontal line from 90 on the cumulative frequency axis to the curve, then read down to the time axis. This gives an estimated 60th percentile of 37 minutes.

(c)

解法一

思路

展开

各组以组中点代表，频数作权重。组距不同不影响加权平均公式，不要误用累计频数作权重。

答题过程

展开

The class midpoints are

7,\ 12,\ 17,\ 24.5,\ 34.5,\ 44.5,\ 54.5,\ 69.5.

Therefore,

\begin{align*} \sum fm=&\,4(7)+11(12)+14(17)+26(24.5)\\[4mm] &\,\hspace{2pt}+48(34.5)+28(44.5)+15(54.5)+4(69.5)\\[4mm] =&\,5032.5. \end{align*}

The estimated mean is

\frac{5032.5}{150}=33.55 \approx\boxed{33.6\text{ minutes}}\quad\text{(3 s.f.)}.

CIE 9709 2026 March Paper 52 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