题目

Problem

Teams of 15 runners took part in a charity run last Saturday. The times taken, in minutes, to complete the course by the runners from the Falcons and the runners from the Kites are shown in the table.

Falcons	38	39	42	44	46	48	50	51	52	56	58	59	64	69	76
Kites	32	40	40	45	47	48	52	54	58	59	59	60	61	63	65

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

(4)

(b) Find the median and the interquartile range of the times for the Falcons.

(3)

Let $x$ and $y$ denote the times, in minutes, of a runner from the Falcons and a runner from the Kites respectively.

It is given that

\sum x=792,\qquad \sum x^2=43\,504,

\sum y=783,\qquad \sum y^2=42\,223.

(3)

题目中文翻译

上周六，每队 15 名跑者参加了一次慈善跑。Falcons 队和 Kites 队各跑者完成全程所用的时间（以分钟为单位）如下表所示。

Falcons	38	39	42	44	46	48	50	51	52	56	58	59	64	69	76
Kites	32	40	40	45	47	48	52	54	58	59	59	60	61	63	65

(a) 画一幅背靠背茎叶图表示这些数据，并将 Falcons 队的数据画在左侧。

(4)

(b) 求 Falcons 队所用时间的中位数和四分位距。

(3)

设 $x$ 和 $y$ 分别表示 Falcons 队和 Kites 队一名跑者所用的时间，单位为分钟。

已知

\sum x=792,\qquad \sum x^2=43\,504,

\sum y=783,\qquad \sum y^2=42\,223.

(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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十位数作茎，个位数作叶。左侧叶从茎向外递增，并写出含两队名称和分钟单位的双侧图例。

答题过程

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Use stems $3,4,5,6,7$ . Place the Falcons on the left and the Kites on the right, with each set of leaves ordered away from the stem.

Key: $1\mid5\mid4$ represents 51 minutes for the Falcons and 54 minutes for the Kites.

(b)

解法一

思路

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15个数据，中位数为第8个；去除中位数后，两半各7个，四分位数分别为第4与第12个。

答题过程

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For the Falcons, the median is the 8th value, so the median is 51 minutes.

The lower quartile is the 4th value, 44 minutes, and the upper quartile is the 12th value, 59 minutes. Thus

\mathrm{IQR}=59-44=\boxed{15\text{ minutes}}.

(c)

解法一

思路

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合并两队时先加总人数、数据总和、平方和，再算总体标准差。

答题过程

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For all 30 runners,

\begin{align*} \sum t=&\,792+783=1575,\\[4mm] \sum t^2=&\,43504+42223=85727. \end{align*}

The mean is

\overline t=\frac{1575}{30}=\boxed{52.5\text{ minutes}}.

The standard deviation is

\begin{align*} \sigma=&\,\sqrt{\frac{85727}{30}-(52.5)^2}\\[4mm] =&\,\sqrt{101.316666\ldots}\\[4mm] =&\,10.065618\ldots\\[4mm] \approx&\,\boxed{10.1\text{ minutes}}. \end{align*}

CIE 9709 2024 November Paper 52 Q6

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