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CIE 9709 2025 November Paper 63 Q4

A Level / CIE / S2

CIE 9709 2025 Nov Paper 63 Paper · Question 4

题目

Problem

The masses of a certain species of animal are known to be normally distributed with standard deviation σ kg\sigma\,\mathrm{kg}. A researcher obtains the masses of a random sample of nn animals of this species and uses these masses to find two confidence intervals (α%\alpha\% and 90%90\%) for the population mean. The width of the α%\alpha\% confidence interval is 1.414×1.414\times the width of the 90%90\% confidence interval.

(a) Find α\alpha.

(3)

(b) Find the probability that the 90%90\% confidence interval contains the population mean given that the α%\alpha\% confidence interval contains the population mean.

(1)
题目中文翻译

已知某种动物的质量服从正态分布,标准差为 σ kg\sigma\,\mathrm{kg}。一名研究人员从该物种随机抽取 nn 只动物,测得它们的质量,并用这些数据求出总体均值的两个置信区间,置信水平分别为 α%\alpha\% 和 90%90\%。α%\alpha\% 置信区间的宽度是 90%90\% 置信区间宽度的 1.4141.414 倍。

(a) 求 α\alpha。

(3)

(b) 已知 α%\alpha\% 置信区间包含总体均值,求 90%90\% 置信区间也包含总体均值的概率。

(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

展开

同一份样本和同一总体标准差使宽度只随正态分位数变化,因此宽度比就是分位数之比。

答题过程

展开

For a confidence interval based on this sample, the width is 2zσ/n2z\sigma/\sqrt n. Using z0.95≈1.645z_{0.95}\approx1.645 for the 90%90\% interval, the given width ratio implies

2zσ/n2(1.645)σ/n= 1.414,z= 1.414(1.645)=2.32603.\begin{align*} \frac{2z\sigma/\sqrt n}{2(1.645)\sigma/\sqrt n}=&\,1.414,\\[4mm] z=&\,1.414(1.645)=2.32603. \end{align*}

Therefore

α100= P(−z<Z<z)= 2Φ(2.32603)−1≈ 0.979983.\begin{align*} \frac\alpha{100}=&\,\mathrm P(-z<Z<z)\\[4mm] =&\,2\Phi(2.32603)-1\\[4mm] \approx&\,0.979983. \end{align*}

Hence α=98.0\boxed{\alpha=98.0} to 33 significant figures.

(b)

解法一

思路

展开

两个区间来自同一样本且中心相同,较窄的 90%90\% 区间包含于较宽区间;利用事件包含关系,而不是假设两个覆盖事件独立。

答题过程

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Both intervals have the same centre, x‾\overline x, and the 90%90\% interval is contained in the wider α%\alpha\% interval. Therefore the event that the 90%90\% interval covers the true mean is a subset of the corresponding event for the wider interval.

Let AA be the event that the 90%90\% interval covers μ\mu, and BB the event that the wider interval covers μ\mu. Since A⊂BA\subset B, using part (a),

P(A∣B)= P(A)P(B)= 0.900.98=4549≈ 0.918.\begin{align*} \mathrm P(A\mid B)=&\,\frac{\mathrm P(A)}{\mathrm P(B)}\\[4mm] =&\,\frac{0.90}{0.98}=\frac{45}{49}\\[4mm] \approx&\,\boxed{0.918}. \end{align*}