题目

Problem

Eric throws darts at a target. The probability that he will hit the centre of the target on one throw is $p$ .

(a) Eric takes $100$ throws and hits the centre $x$ times. He uses this result to find an approximate $95\%$ confidence interval for $p$ . The width of this interval is $0.165$ correct to $3$ significant figures.

Find the two possible values of $x$ .

(4)

Eric takes some more throws. He uses his results to calculate two approximate $\alpha\%$ confidence intervals for $p$ (where $\alpha>90$ ).

It is given that the probability that exactly one of these two intervals contains the true value of $p$ is $0.1128$ .

(b) Find the value of $\alpha$ .

(3)

题目中文翻译

Eric 向靶子投飞镖。他一次投掷命中靶心的概率为 $p$ 。

(a) Eric 投掷 $100$ 次，命中靶心 $x$ 次。他用这一结果求出 $p$ 的一个近似 $95\%$ 置信区间。这个区间的宽度为 $0.165$ （保留 $3$ 位有效数字）。

求 $x$ 的两个可能值。

(4)

Eric 又投掷了若干次。他用所得结果计算出 $p$ 的两个近似 $\alpha\%$ 置信区间（其中 $\alpha>90$ ）。

已知这两个区间中恰有一个包含 $p$ 的真实值的概率为 $0.1128$ 。

(b) 求 $\alpha$ 的值。

(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)给出两个区间恰有一个包含真实比例的概率，但未说明两个区间的覆盖事件独立。

实际上应为：仅凭所给概率不能唯一确定置信水平；还需说明两区间如何抽样以及覆盖事件之间的关系。在额外假设覆盖事件独立时，才得到评分标准的 $\alpha=94$ 。

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

展开

置信区间的宽度是两倍误差界。令样本比例为 $x/100$ ，解出两个对称的比例，再用命中次数必须为整数和题目给出的四舍五入精度核对。

答题过程

展开

Put $\widehat p=x/100$ . An approximate $95\%$ confidence interval is

\widehat p\pm1.96\sqrt{\frac{\widehat p(1-\widehat p)}{100}}.

Its width is approximately $0.165$ , so

\begin{align*} 2(1.96)\sqrt{\frac{\widehat p(1-\widehat p)}{100}} \approx&\,0.165,\\[4mm] \widehat p(1-\widehat p) \approx&\,100\bigg(\frac{0.165}{3.92}\bigg)^2\\[4mm] \approx&\,0.177172272. \end{align*}

Solving $\widehat p^2-\widehat p+0.177172272=0$ gives $\widehat p\approx0.230134$ or $0.769866$ , corresponding to $x\approx23.0134$ or $76.9866$ .

As $x$ is an integer, the candidates are $23$ and $77$ . Each gives the width

3.92\sqrt{\frac{0.23(0.77)}{100}}\approx0.164966,

which rounds to $0.165$ to $3$ significant figures. Adjacent integer counts give widths outside the rounding interval $[0.1645,0.1655)$ ; the width is monotone on each side of $x=50$ . Therefore

\boxed{x=23\quad\text{or}\quad x=77}.

CIE 9709 2026 June Paper 62 Q5

题目

解答

(a)

解法一

思路

答题过程