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IAL 2021 June S3 Q3

A Level / Edexcel / S3

IAL 2021 June Paper · Question 3

题目

Problem

Components are manufactured such that their length in mm is normally distributed with mean μ\mu and variance σ2\sigma^2. Below is a 95% confidence interval for μ\mu calculated from a random sample of components.

(11.52, 13.75)(11.52,\ 13.75)

Using the same random sample,

(a) find a 90% confidence interval for μ\mu.

(6)

Four 90% confidence intervals are found from independent random samples.

(b) Calculate the probability that only 3 of these 4 intervals will contain μ\mu.

(2)
(Total for Question 3 is 8 marks)
题目中文翻译

元件的长度（单位 mm）服从均值为 μ\mu、方差为 σ2\sigma^2 的正态分布。下面是根据一个元件样本算出的 μ\mu 的 95% 置信区间。

(11.52, 13.75)(11.52,\ 13.75)

使用同一个样本，

(a) 求 μ\mu 的 90% 置信区间。

(b) 从相互独立的 4 个随机样本中得到 4 个 90% 置信区间。求其中恰好只有 3 个区间包含 μ\mu 的概率。

解答

(a)

解法一

思路

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同一个样本给出的不同置信水平区间具有相同的样本均值与标准误。先由原 95%95\% 区间的中点求 xˉ\bar{x}，再用半宽除以 1.961.96 求标准误；最后改用 90%90\% 区间的临界值 1.64491.6449。

答题过程

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The sample mean is the midpoint of the 95%95\% confidence interval:

xˉ= 11.52+13.752= 12.635.\begin{align*} \bar{x} =&\,\frac{11.52+13.75}{2}\\ =&\,12.635. \end{align*}

The half-width of the 95%95\% confidence interval is

13.75−12.635=1.115.13.75-12.635=1.115.

Since the 95%95\% critical value is 1.961.96, the standard error is

SE⁡= 1.1151.96= 0.568877…\begin{align*} \operatorname{SE} =&\,\frac{1.115}{1.96}\\ =&\,0.568877\ldots \end{align*}

For a 90%90\% confidence interval, the critical value is 1.64491.6449. Hence the interval is

 xˉ±1.6449SE⁡= 12.635±1.6449(0.568877…)= 12.635±0.9357…= (11.699…, 13.570…).\begin{align*} &\,\bar{x}\pm1.6449\operatorname{SE}\\ =&\,12.635\pm1.6449(0.568877\ldots)\\ =&\,12.635\pm0.9357\ldots\\ =&\,(11.699\ldots,\ 13.570\ldots). \end{align*}

Therefore, to 33 significant figures, the 90%90\% confidence interval is

(11.7, 13.6) mm.\boxed{(11.7,\ 13.6)\text{ mm}}.

(b)

解法一

思路

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每个 90%90\% 置信区间包含 μ\mu 的概率为 0.90.9，不包含的概率为 0.10.1。四个区间来自独立样本，因此包含 μ\mu 的区间数服从 Bin⁡(4,0.9)\operatorname{Bin}(4,0.9)，计算恰好成功三次的概率。

答题过程

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Let XX be the number of the four confidence intervals that contain μ\mu. Then

X∼Bin⁡(4,0.9).X\sim\operatorname{Bin}(4,0.9).

Therefore,

P⁡(X=3)= (43)(0.9)3(0.1)= 0.2916≈ 0.292.\begin{align*} \operatorname{P}(X=3) =&\,\binom{4}{3}(0.9)^3(0.1)\\ =&\,0.2916\\ \approx&\,\boxed{0.292}. \end{align*}