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CIE 9709 2026 June Paper 63 Q1

A Level / CIE / S2

CIE 9709 2026 June Paper 63 Paper · Question 1

题目

Problem

A five-sided spinner has edges marked 11, 22, 33, 44, 55. Ayesha believes that the spinner is biased so that it lands on five more often than it should. She spins the spinner 200200 times, and it lands on five on 6060 spins.

Use Ayesha’s result to calculate an approximate 94%94\% confidence interval for the probability that the spinner lands on five on one spin.

(3)
题目中文翻译

一个五面陀螺的各边分别标有 11、22、33、44、55。Ayesha 认为这个陀螺不均匀，落在数字 55 的频率高于应有水平。她将陀螺转动 200200 次，其中有 6060 次落在数字 55。

用 Ayesha 的结果，计算陀螺转动一次落在数字 55 的概率的一个近似 94%94\% 置信区间。

(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

解法一

思路

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用样本比例估计真实概率。置信度 94%94\% 对应两尾各 3%3\%，所以使用正态累计概率 0.970.97 的分位数。

答题过程

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The sample proportion is

p^=60200=0.3.\widehat p=\frac{60}{200}=0.3.

For a 94%94\% confidence interval, z=Φ−1(0.97)≈1.88079z=\Phi^{-1}(0.97)\approx1.88079. Therefore the approximate interval is

 p^±zp^(1−p^)200= 0.3±1.880790.3(0.7)200≈ 0.3±0.06095.\begin{align*} &\,\widehat p\pm z\sqrt{\frac{\widehat p(1-\widehat p)}{200}}\\[4mm] =&\,0.3\pm1.88079\sqrt{\frac{0.3(0.7)}{200}}\\[4mm] \approx&\,0.3\pm0.06095. \end{align*}

Hence the required interval is (0.239, 0.361)\boxed{(0.239,\ 0.361)} to 33 significant figures.