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CIE 9709 2023 November Paper 51 Q1

A Level / CIE / S1

CIE 9709 2023 Nov Paper 51 Paper · Question 1

题目

Problem

The times taken by 120 children to complete a particular puzzle are represented in the cumulative frequency graph.

(a) Use the graph to estimate the interquartile range of the data.

(2)

35%35\% of the children took longer than TT seconds to complete the puzzle.

(b) Use the graph to estimate the value of TT.

(2)
题目中文翻译

累积频数图表示了 120 名儿童完成一道特定谜题所用的时间。

(a) 利用该图估计这些数据的四分位距。

(2)

有 35%35\% 的儿童完成谜题所用的时间超过 TT 秒。

(b) 利用该图估计 TT 的值。

(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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120人的两个四分位数对应累计频数30与90,从曲线读时间后相减。

答题过程

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Read from cumulative frequencies 1204=30\frac{120}{4}=30 and 3(120)4=90\frac{3(120)}4=90 to the curve, then down to the time axis. This gives

Q1≈23.7 s,Q3≈31 s.Q_1\approx23.7\text{ s},\qquad Q_3\approx31\text{ s}.

Hence the estimated interquartile range is

31−23.7=7.3 seconds.31-23.7=\boxed{7.3\text{ seconds}}.

(b)

解法一

思路

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35%用时超过T,故65%用时不超过T,对应累计频数78。

答题过程

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The required cumulative frequency is

120(1−0.35)=78.120(1-0.35)=78.

Read horizontally from 78 to the curve and then down to the time axis. This gives T≈28.5T\approx28.5 seconds.