题目

Problem

Last Sunday, teams of runners took part in a charity event. The time taken, in seconds, to run $50\,\mathrm{m}$ was recorded, correct to 1 decimal place, for each runner. The times recorded for 11 runners from each of the Gulls and the Herons are shown in the table.

Gulls	7.9	8.2	8.3	8.6	8.6	8.8	9.2	9.7	9.8	10.0	10.4
Herons	9.5	9.9	8.5	8.1	9.2	10.8	8.3	9.7	9.3	9.9	8.7

(a) Draw a back-to-back stem-and-leaf diagram to represent this information, with Gulls on the left-hand side.

(4)

(b) Find the median and the interquartile range of the times of the runners from the Gulls.

(3)

Two other teams of runners, the Eagles and the Swifts, also took part in the event. The recorded times in seconds for 20 runners from the Eagles and 30 runners from the Swifts are denoted by $x$ and $y$ respectively.

It is given that $\sum x = 175.0$ and that the mean of $y$ is $8.4$ .

(2)

It is given that $\sum x^2 = 1823.0$ .

It is also known that the standard deviation of the times taken by all 50 runners is $1.38$ seconds.

(d) Find the value of $\sum y^2$ , correct to 1 decimal place.

(3)

题目中文翻译

上星期日，若干跑步队参加了一项慈善活动。记录每名跑步者跑完 $50\,\mathrm{m}$ 所用的时间，以秒为单位，精确到小数点后 1 位。下表列出了 Gulls 队和 Herons 队各 11 名跑步者的用时。

Gulls	7.9	8.2	8.3	8.6	8.6	8.8	9.2	9.7	9.8	10.0	10.4
Herons	9.5	9.9	8.5	8.1	9.2	10.8	8.3	9.7	9.3	9.9	8.7

(a) 画出背靠背茎叶图来表示这些数据，将 Gulls 队的数据放在左侧。

(4)

(b) 求 Gulls 队跑步者用时的中位数和四分位距。

(3)

另外两支跑步队 Eagles 和 Swifts 也参加了这项活动。Eagles 队 20 名跑步者和 Swifts 队 30 名跑步者记录的用时（单位为秒）分别用 $x$ 和 $y$ 表示。

已知 $\sum x = 175.0$ ，且 $y$ 的均值为 $8.4$ 。

(2)

已知 $\sum x^2 = 1823.0$ 。

还知道全部 50 名跑步者用时的标准差为 $1.38$ 秒。

(d) 求 $\sum y^2$ 的值，精确到小数点后 1 位。

(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。

题目勘误

官方原题：(c)(d)给出Eagles有20人、 $\sum x=175.0$ 、 $\sum x^2=1823.0$ ，Swifts有30人、均值为 $8.4$ ，并称全部50人的标准差为 $1.38$ 秒。

实际上应为：这些统计条件不能同时成立。前述数据要求全部50人的标准差至少约为 $2.42$ 秒，不能等于 $1.38$ 秒。尚未找到官方明确更正，不能唯一确定应改哪项数据。

本页保留官方英文及忠实翻译，未替换原题条件。

解答

(a)

解法一

思路

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以整数部分为茎、十分位为叶，Gulls在左侧。左边由茎向外递增，保留两个8.6等重复数据。

答题过程

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Use the integer part as the stem and the tenths digit as each leaf, with Gulls on the left.

Key: $7\,|\,9\,|\,5$ means 9.7 seconds for Gulls and 9.5 seconds for Herons.

(b)

解法一

思路

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11个有序数据，中位数为第6个，下、上四分位数为第3及第9个。

答题过程

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For Gulls, the median is the 6th value, 8.8 seconds.

The lower quartile is 8.3 seconds and the upper quartile is 9.8 seconds. Thus

\text{interquartile range}=9.8-8.3 =\boxed{1.5\text{ seconds}}.

(c)

解法一

思路

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Swifts有30人，均值8.4，所以先还原这组的总时间，再加Eagles的总时间，除以50。

答题过程

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The total time for the Swifts is $30(8.4)=252.0$ seconds. Therefore the mean for all 50 runners is

\frac{175.0+252.0}{20+30} =\frac{427.0}{50} =\boxed{8.54\text{ seconds}}.

CIE 9709 2025 June Paper 51 Q3

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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答题过程