题目

Problem

A baker produces bread buns and bread rolls. The weights of buns, $B$ grams, and the weights of rolls, $R$ grams, are such that $B \sim N(55, 1.3^2)$ and $R \sim N(51, 1.2^2)$ .

A bun and a roll are selected at random.

(a) Find the probability that the bun weighs less than 110% of the weight of the roll.

(5)

Two buns are chosen at random.

(b) Find the probability that their weights differ by more than 1 gram.

(6)

The baker sells bread in bags. Each bag contains either 10 buns or 11 rolls. The weight of an empty bag, $S$ grams, is such that $S \sim N(3, 0.2^2)$

(6)

(Total for Question 6 is 17 marks)

题目中文翻译

一位面包师生产面包小圆包和面包卷。小圆包的重量 $B$ 克和面包卷的重量 $R$ 克分别服从 $B \sim N(55, 1.3^2)$ 和 $R \sim N(51, 1.2^2)$ 。

随机选取一个小圆包和一个面包卷。

(a) 求小圆包的重量小于面包卷重量的 110% 的概率。

(5)

随机选取两个小圆包。

(b) 求它们重量相差超过 1 克的概率。

(6)

这位面包师把面包装袋出售。每袋要么装 10 个小圆包，要么装 11 个面包卷。空袋重量 $S$ 克，且 $S \sim N(3, 0.2^2)$ 。

(6)

解答

(a)

解法一

思路

展开

把“不足面包卷重量的 $110\%$ ”写成 $B<1.1R$ 。令 $W=B-1.1R$ ，原事件便成为 $W<0$ 。由于随机抽取的面包小圆包与面包卷相互独立， $W$ 仍服从正态分布，其方差是两个方差相加。

答题过程

展开

Let

W=B-1.1R.

Since $B$ and $R$ are independent,

\begin{align*} E(W)=&\,55-1.1(51)\\ =&\,-1.1, \end{align*}

and

\begin{align*} \operatorname{Var}(W)=&\,1.3^2+1.1^2(1.2^2)\\ =&\,3.4324. \end{align*}

Therefore,

W\sim N(-1.1,3.4324).

Hence

\begin{align*} P(B<1.1R)=&\,P(W<0)\\ =&\,P\bigg(Z<\frac{0-(-1.1)}{\sqrt{3.4324}}\bigg)\\ =&\,P(Z<0.5937\ldots)\\ =&\,0.723655\ldots \end{align*}

Thus, to 3 significant figures,

\boxed{P(B<1.1R)=0.724}.

(b)

解法一

思路

展开

用 $B_1$ 、 $B_2$ 表示两只独立抽取的小圆包重量，并令 $X=B_1-B_2$ 。两者重量相差超过 $1$ 克即 $|X|>1$ 。由于 $X$ 的分布关于 $0$ 对称，可只求右侧尾部 $P(X>1)$ ，再乘以 $2$ 。

答题过程

展开

Let $B_1$ and $B_2$ be the weights of the two independently selected buns, and let

X=B_1-B_2.

Then

\begin{align*} E(X)=&\,55-55=0,\\ \operatorname{Var}(X)=&\,1.3^2+1.3^2=3.38. \end{align*}

Therefore,

X\sim N(0,3.38).

By symmetry,

\begin{align*} P(|X|>1)=&\,2P(X>1)\\ =&\,2P\bigg(Z>\frac{1}{\sqrt{3.38}}\bigg)\\ =&\,2P(Z>0.5439\ldots)\\ =&\,0.58649\ldots \end{align*}

Thus, to 3 significant figures, the required probability is

\boxed{0.586}.

(c)

解法一

思路

展开

分别设一袋小圆包和一袋面包卷的总重量为 $V$ 、 $Y$ 。每袋都包含一个独立的空袋重量，因此比较 $Y-V$ 时，两个空袋重量的方差都要计入。先求 $V$ 与 $Y$ 的均值和方差，再令 $D=Y-V$ ；所求事件就是 $D>0$ 。

答题过程

展开

Let $S_B$ and $S_R$ be the independent weights of the empty bags used for the buns and the rolls respectively. Let

V=B_1+B_2+\cdots+B_{10}+S_B

be the total weight of a bag of buns, and let

Y=R_1+R_2+\cdots+R_{11}+S_R

be the total weight of a bag of rolls.

For the bag of buns,

\begin{align*} E(V)=&\,10(55)+3=553,\\ \operatorname{Var}(V)=&\,10(1.3^2)+0.2^2=16.94. \end{align*}

Hence

V\sim N(553,16.94).

For the bag of rolls,

\begin{align*} E(Y)=&\,11(51)+3=564,\\ \operatorname{Var}(Y)=&\,11(1.2^2)+0.2^2=15.88. \end{align*}

Hence

Y\sim N(564,15.88).

Now let

D=Y-V.

Since the two bags and all the items selected are independent,

\begin{align*} E(D)=&\,564-553=11,\\ \operatorname{Var}(D)=&\,15.88+16.94=32.82. \end{align*}

Therefore,

D\sim N(11,32.82).

The required probability is

\begin{align*} P(V<Y)=&\,P(D>0)\\ =&\,P\bigg(Z>\frac{0-11}{\sqrt{32.82}}\bigg)\\ =&\,P(Z>-1.920\ldots)\\ =&\,0.972577\ldots \end{align*}

Thus, to 3 significant figures,

\boxed{P(V<Y)=0.973}.

S3 2021 June Q6

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