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CIE 9709 2026 June Paper 65 Q3

A Level / CIE / S2

CIE 9709 2026 June Paper 65 Paper · Question 3

题目

Problem

A researcher chose a random sample of 150150 adults in Dombridge for a survey about dog ownership.

The number of adults in the survey who owned a dog was aa. An approximate 94%94\% confidence interval for the population proportion, pp, of all the adults in Dombridge who owned a dog was calculated. The width of the interval was 0.132410.13241, correct to 55 decimal places.

(a) Calculate the two possible values of aa.

(5)

Two approximate 94%94\% confidence intervals were found.

(b) Given that at least one of these intervals contain the true value of pp, find the probability that both intervals contain the true value of pp.

(3)

The survey included a question about the distance, XX kilometres, that the adults walked with their dog each day. The results were summarised as follows.

n=150∑x=462∑x2=1485n=150\qquad \sum x=462\qquad \sum x^2=1485

(c) Find unbiased estimates of the mean and variance of XX.

(3)
题目中文翻译

一名研究人员从 Dombridge 随机抽取 150150 名成年人，就养狗情况进行调查。

受访成年人中有 aa 人养狗。用 pp 表示 Dombridge 所有成年人中养狗者的总体比例，并求得 pp 的一个近似 94%94\% 置信区间。这个区间的宽度为 0.132410.13241（保留 55 位小数）。

(a) 计算 aa 的两个可能值。

(5)

现求得两个近似 94%94\% 置信区间。

(b) 已知这两个区间中至少有一个包含 pp 的真实值，求两个区间都包含 pp 的真实值的概率。

(3)

调查中有一道问题，询问这些成年人每天遛狗走过的距离 XX（以千米为单位）。结果汇总如下。

n=150∑x=462∑x2=1485n=150\qquad \sum x=462\qquad \sum x^2=1485

(c) 求 XX 的均值和方差的无偏估计值。

(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。

题目勘误

官方原题：(b)只说明找到两个近似 94%94\% 置信区间，未说明它们的覆盖事件是否独立。

实际上应为：同一置信水平只给出各自的覆盖概率，不能确定两者的联合概率，因此原条件不足以唯一计算所求条件概率。额外假设覆盖事件独立时，才得到评分标准的 47/5347/53。

本页保留官方英文与忠实翻译，未自行补入独立条件。

解答

(a)

解法一

思路

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使用 94%94\% 区间的标准正态表值 1.8811.881，由宽度反推样本比例。两根关于 1/21/2 对称，最后回到整数人数并核对给定五位小数。

答题过程

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Let p^=a/150\widehat p=a/150. Using the normal-table value z=1.881z=1.881 for a 94%94\% confidence interval, its width gives

2(1.881)p^(1−p^)150≈0.13241.2(1.881)\sqrt{\frac{\widehat p(1-\widehat p)}{150}}\approx0.13241.

Squaring and rearranging,

p^(1−p^)≈ 150(0.132413.762)2≈ 0.1858212,p^2−p^+0.1858212≈ 0.\begin{align*} \widehat p(1-\widehat p) \approx&\,150\bigg(\frac{0.13241}{3.762}\bigg)^2\\[4mm] \approx&\,0.1858212,\\[4mm] \widehat p^2-\widehat p+0.1858212\approx&\,0. \end{align*}

Thus p^≈0.246665\widehat p\approx0.246665 or 0.7533350.753335, giving a≈36.9997a\approx36.9997 or 113.0003113.0003. The integer possibilities are therefore

a=37ora=113.\boxed{a=37\quad\text{or}\quad a=113}.

For either value, the width calculated with z=1.881z=1.881 is

2(1.881)(37/150)(113/150)150≈0.1324104,2(1.881)\sqrt{\frac{(37/150)(113/150)}{150}} \approx0.1324104,

which rounds to the given 0.132410.13241 to 55 decimal places. Adjacent integer values give widths outside its rounding interval.

(c)

解法一

思路

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使用给定的全体 150150 个观测值计算无偏均值与方差，方差分母为 149149。

答题过程

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The unbiased estimate of the mean is

x‾=462150=3.08 km.\boxed{\overline x=\frac{462}{150}=3.08\ \text{km}}.

The unbiased estimate of the variance is

s2= 1149(1485−4622150)= 1485−1422.96149= 62.04149≈ 0.416 km2.\begin{align*} s^2=&\,\frac1{149}\bigg(1485-\frac{462^2}{150}\bigg)\\[4mm] =&\,\frac{1485-1422.96}{149}\\[4mm] =&\,\frac{62.04}{149}\\[4mm] \approx&\,\boxed{0.416\ \text{km}^2}. \end{align*}