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CIE 9709 2026 June Paper 52 Q6

A Level / CIE / S1

CIE 9709 2026 June Paper 52 Paper · Question 6

题目

Problem

The heights, in cm, of the 15 rugby players in each of two teams, the Hippos and the Zebras, are shown in the following table.

Hippos165168172172176178181184185188188190191192195
Zebras183192172167184189171194191179186182177175186

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

(4)

(b) Find the interquartile range of the heights of the players in the Hippos team.

(2)

The player in the Hippos with height 184 cm184\,\mathrm{cm} is injured. A new player replaces him, and this causes the mean height of the 15 players in the team to increase by 0.5 cm0.5\,\mathrm{cm}.

(c) Find the height of the new player.

(2)
题目中文翻译

Hippos 和 Zebras 两支橄榄球队各有 15 名球员,他们的身高如下表所示,单位为厘米。

Hippos165168172172176178181184185188188190191192195
Zebras183192172167184189171194191179186182177175186

(a) 画出背靠背茎叶图表示这些数据,将 Hippos 放在左侧。

(4)

(b) 求 Hippos 队球员身高的四分位距。

(2)

Hippos 队中身高为 184 cm184\,\mathrm{cm} 的球员受伤。一名新球员替换了他,使全队 15 名球员的平均身高增加了 0.5 cm0.5\,\mathrm{cm}。

(c) 求新球员的身高。

(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

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将前两位作茎,个位作叶。Hippos放左侧并由茎向外递增,Zebras放右侧;保留所有重复值,并加上带队名和单位的非对称图例。

答题过程

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Use the first two digits as stems and the final digit as each leaf, with Hippos on the left.

Key: 2 ∣ 17 ∣ 92\,|\,17\,|\,9 represents 172 cm172\,\mathrm{cm} for Hippos and 179 cm179\,\mathrm{cm} for Zebras.

(b)

解法一

思路

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15个数据的中位数在第8位。下、上四分位数分别是去掉中位数后的两组中位数,即原有序数据的第4及第12位。

答题过程

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The Hippos data are already in ascending order. The lower quartile is the 4th value, 172 cm172\,\mathrm{cm}, and the upper quartile is the 12th value, 190 cm190\,\mathrm{cm}. Hence

interquartile range=190−172=18 cm.\text{interquartile range}=190-172=\boxed{18\,\mathrm{cm}}.

(c)

解法一

思路

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人数保持15人,均值增加0.5,说明总身高增加7.5厘米。只有被替换的球员改变,因此新球员比原球员高7.5厘米。

答题过程

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The increase in the total height is

15×0.5=7.5 cm.15\times0.5=7.5\,\mathrm{cm}.

Only one player has changed, so the new player’s height is

184+7.5=191.5 cm.184+7.5=\boxed{191.5\,\mathrm{cm}}.