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CIE 9709 2025 June Paper 65 Q2

A Level / CIE / S2

CIE 9709 2025 June Paper 65 Paper · Question 2

题目

Problem

A random sample of 200200 values of a random variable XX gives the following results.

n=200∑(x−2)=60∑(x−2)2=20n=200\qquad \sum(x-2)=60\qquad \sum(x-2)^2=20

(a) Find a 95%95\% confidence interval for the population mean of XX.

(6)

(b) State which part of your solution to part (a) makes use of the Central Limit Theorem.

(1)
题目中文翻译

从随机变量 XX 的取值中随机抽取 200200 个,得到下列结果。

n=200∑(x−2)=60∑(x−2)2=20n=200\qquad \sum(x-2)=60\qquad \sum(x-2)^2=20

(a) 求 XX 的总体均值的一个 95%95\% 置信区间。

(6)

(b) 指出你在 (a) 的解答中哪一步用到了中心极限定理。

(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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平移不改变方差,均值则需要加回二;无偏方差用分母一百九十九。

答题过程

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Put Y=X−2Y=X-2. Its sample mean is 60/200=0.360/200=0.3, so x‾=2.3\overline x=2.3. Since translation does not change variance, the unbiased variance estimate for XX is

s2= 1199(20−602200)= 2199.\begin{align*} s^2=&\,\frac1{199}\bigg(20-\frac{60^2}{200}\bigg)\\[4mm] =&\,\frac2{199}. \end{align*}

By the Central Limit Theorem, the sample mean is approximately normal for the large random sample. An approximate 95%95\% confidence interval is

 x‾±1.96s2200= 2.3±1.962/199200≈ 2.3±0.013894.\begin{align*} &\,\overline x\pm1.96\sqrt{\frac{s^2}{200}}\\[4mm] =&\,2.3\pm1.96\sqrt{\frac{2/199}{200}}\\[4mm] \approx&\,2.3\pm0.013894. \end{align*}

Hence the interval is (2.29, 2.31)\boxed{(2.29,\ 2.31)} to 33 significant figures.

(b)

解法一

思路

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明确指出使用正态临界值的步骤,不只说总体未给正态。

答题过程

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The Central Limit Theorem justifies the normal approximation for the sample mean, and hence the use of the normal critical value z=1.96\boxed{z=1.96} in the confidence-interval formula.