题目

Problem

A company produces bricks.

The weight of a brick, $B$ kg, is such that $B \sim N(1.96, (\sqrt{0.003})^2)$ .

Two bricks are chosen at random.

(a) Find the probability that the difference in weight of the 2 bricks is greater than 0.1 kg

(5)

A random sample of $n$ bricks is to be taken.

(b) Find the minimum sample size such that the probability of the sample mean being greater than 2 is less than 1%

(5)

The bricks are randomly selected and stacked on pallets.

The weight of an empty pallet, $E$ kg, is such that $E \sim N(21.8, (\sqrt{0.6})^2)$ .

The random variable $M$ represents the total weight of a pallet stacked with 500 bricks.

The random variable $T$ represents the total weight of a container of cement.

Given that $T$ is independent of $M$ and that $T \sim N(774, (\sqrt{1.8})^2)$

(7)

(Total 17 marks)

题目中文翻译

一家公司生产砖块。

砖块重量 $B$ kg 服从 $B \sim N(1.96, (\sqrt{0.003})^2)$ 。

随机选取 2 块砖。

(a) 求这 2 块砖重量差大于 0.1 kg 的概率。

随机抽取 n 块砖组成样本。

(b) 求最小样本量，使样本均值大于 2 的概率小于 1%。

砖块被随机选取并堆放在托盘上。

空托盘的重量 $E$ kg 服从 $E \sim N(21.8, (\sqrt{0.6})^2)$ 。

随机变量 $M$ 表示装有 500 块砖的托盘总重量。

随机变量 $T$ 表示一箱水泥的总重量。

已知 $T$ 与 $M$ 独立，且 $T \sim N(774, (\sqrt{1.8})^2)$ 。

解答

(a)

解法一

思路

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令 $D=B_1-B_2$ 。两块砖独立且同分布，所以 $D$ 的均值为 $0$ ，方差为两个方差之和 $0.006$ 。“重量差大于 $0.1$ ”包括 $D>0.1$ 与 $D<-0.1$ 两种对称情形，因此需把单侧概率乘以 $2$ 。

答题过程

展开

Let

D=B_1-B_2.

Since the two bricks are selected independently,

\begin{align*} \operatorname{E}(D) =&\,1.96-1.96\\ =&\,0, \end{align*}

and

\begin{align*} \operatorname{Var}(D) =&\,0.003+0.003\\ =&\,0.006. \end{align*}

Thus $D\sim N(0,0.006)$ . By symmetry,

\begin{align*} \operatorname{P}(|B_1-B_2|>0.1) =&\,2\operatorname{P}(D>0.1)\\ =&\,2\operatorname{P}\left( Z>\frac{0.1}{\sqrt{0.006}} \right)\\ =&\,2\operatorname{P}(Z>1.2909\ldots)\\ =&\,0.1967\ldots\\ \approx&\,\boxed{0.197}. \end{align*}

(b)

解法一

思路

展开

先写出样本均值的分布，其方差为总体方差除以 $n$ 。要使右尾概率小于 $0.01$ ，数值 $2$ 标准化后的 $z$ 值必须大于标准正态分布第 $99$ 百分位数 $2.3263$ 。解出 $n$ 的严格下界后取最小整数。

答题过程

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The sample mean has distribution

\overline{B}\sim N\left(1.96,\frac{0.003}{n}\right).

Therefore,

\operatorname{P}(\overline{B}>2) =\operatorname{P}\left( Z>\frac{2-1.96}{\sqrt{0.003/n}} \right).

For this probability to be less than $0.01$ ,

\frac{2-1.96}{\sqrt{0.003/n}}>2.3263.

Hence

\begin{align*} \frac{0.04\sqrt{n}}{\sqrt{0.003}} >&\,2.3263\\ \sqrt{n} >&\,\frac{2.3263\sqrt{0.003}}{0.04}\\ n >&\,10.146\ldots \end{align*}

Therefore, the minimum sample size is

\boxed{n=11}.

(c)

解法一

思路

展开

先把托盘总重写成空托盘与 $500$ 块独立砖重之和，求出 $M$ 的均值和方差。再令 $X=4T-3M$ ，把目标事件改写成 $X>100$ 。独立正态变量的线性组合仍为正态分布，计算方差时系数要平方。

答题过程

展开

The total pallet weight is

M=E+\sum_{i=1}^{500}B_i.

Therefore,

\begin{align*} \operatorname{E}(M) =&\,21.8+500(1.96)\\ =&\,1001.8, \end{align*}

and

\begin{align*} \operatorname{Var}(M) =&\,0.6+500(0.003)\\ =&\,2.1. \end{align*}

Hence

M\sim N(1001.8,2.1).

Let

X=4T-3M.

Since $T$ and $M$ are independent,

\begin{align*} \operatorname{E}(X) =&\,4(774)-3(1001.8)\\ =&\,90.6, \end{align*}

and

\begin{align*} \operatorname{Var}(X) =&\,4^2(1.8)+(-3)^2(2.1)\\ =&\,47.7. \end{align*}

Thus $X\sim N(90.6,47.7)$ , and

\begin{align*} \operatorname{P}(4T>100+3M) =&\,\operatorname{P}(X>100)\\ =&\,\operatorname{P}\left( Z>\frac{100-90.6}{\sqrt{47.7}} \right)\\ =&\,\operatorname{P}(Z>1.3610\ldots)\\ =&\,0.08675\ldots\\ \approx&\,\boxed{0.0868}. \end{align*}

IAL 2021 Oct S3 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程