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CIE 9709 2025 November Paper 62 Q4

A Level / CIE / S2

CIE 9709 2025 Nov Paper 62 Paper · Question 4

题目

Problem

An inspector believes that 18%18\% of cups made at a certain factory contain flaws. The factory owner claims that the true percentage is less than 18%18\%. The inspector examines a random sample of 4040 cups and finds that 33 of them contain flaws.

(a) Stating a necessary assumption, use a binomial distribution to test the factory owner’s claim at the 5%5\% significance level.

(6)

(b) Explain why it would not be appropriate to use the Poisson approximation to the binomial distribution to carry out the test in part (a).

(1)
题目中文翻译

一名检验员认为，某工厂生产的杯子中有 18%18\% 存在瑕疵。厂主声称真实比例低于 18%18\%。检验员随机抽取 4040 个杯子进行检查，发现其中有 33 个存在瑕疵。

(a) 写出一个必要的假设，并用二项分布在 5%5\% 的显著性水平下检验厂主的说法。

(6)

(b) 说明为什么在 (a) 的检验中不宜用泊松分布近似二项分布。

(1)
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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明确杯子缺陷之间独立且缺陷概率相同，再用精确二项左尾概率进行检验；不能把恰好三个的概率当作检验概率。

答题过程

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Assume that flaws occur independently from cup to cup and that the probability of a flaw is the same for every cup. Let pp be the population proportion of flawed cups. Test

H0:p=0.18,H1:p<0.18.H_0:p=0.18,\qquad H_1:p<0.18.

Under H0H_0, the number KK of flawed cups in the sample has distribution B(40,0.18)\mathrm B(40,0.18). For the observation K=3K=3,

P(K≤3)= (0.82)40+40(0.18)(0.82)39 +(402)(0.18)2(0.82)38 +(403)(0.18)3(0.82)37≈ 0.0542.\begin{align*} \mathrm P(K\le3) =&\,(0.82)^{40}+40(0.18)(0.82)^{39}\\[4mm] &\,\hspace{2pt}+\binom{40}{2}(0.18)^2(0.82)^{38}\\[4mm] &\,\hspace{4pt}+\binom{40}{3}(0.18)^3(0.82)^{37}\\[4mm] \approx&\,0.0542. \end{align*}

Since 0.0542>0.050.0542>0.05, do not reject H0H_0. There is insufficient evidence at the 5%5\% significance level that the population proportion of flawed cups is less than 18%18\%, so the owner’s claim is not supported.

(b)

解法一

思路

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Poisson 近似需要二项成功概率很小，本题的 0.180.18 不符合这一条件。

答题过程

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The probability p=0.18\boxed{p=0.18} is not small enough for a suitable Poisson approximation to the binomial distribution.