CalcGospel 國際數學圖譜
返回

IAL 2021 June S3 Q6

A Level / Edexcel / S3

IAL 2021 June Paper · Question 6

题目

Problem

A baker produces bread buns and bread rolls. The weights of buns, BB grams, and the weights of rolls, RR grams, are such that B∼N(55,1.32)B \sim N(55, 1.3^2) and R∼N(51,1.22)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, SS grams, is such that S∼N(3,0.22)S \sim N(3, 0.2^2)

(c) Find the probability that a bag of buns weighs less than a bag of rolls.

(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 BB 克和面包卷的重量 RR 克分别服从 B∼N(55,1.32)B \sim N(55, 1.3^2) 和 R∼N(51,1.22)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。

随机选取两个小圆包。

(b) 求它们重量相差超过 1 克的概率。

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 SS 克，且 S∼N(3,0.22)S \sim N(3, 0.2^2)。

(c) 求一袋小圆包的总重量小于一袋面包卷的总重量的概率。

解答

(a)

解法一

思路

展开

把“不足面包卷重量的 110%110\%”写成 B<1.1RB<1.1R。令 W=B−1.1RW=B-1.1R，原事件便成为 W<0W<0。由于随机抽取的面包小圆包与面包卷相互独立，WW 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.W=B-1.1R.

Since BB and RR are independent,

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

and

Var⁡(W)= 1.32+1.12(1.22)= 3.4324.\begin{align*} \operatorname{Var}(W)=&\,1.3^2+1.1^2(1.2^2)\\ =&\,3.4324. \end{align*}

Therefore,

W∼N(−1.1,3.4324).W\sim N(-1.1,3.4324).

Hence

P(B<1.1R)= P(W<0)= P(Z<0−(−1.1)3.4324)= P(Z<0.5937…)= 0.7237…\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)\\ =&\,0.7237\ldots \end{align*}

Thus, to 3 significant figures,

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

(b)

解法一

思路

展开

用 B1B_1、B2B_2 表示两只独立抽取的小圆包重量，并令 X=B1−B2X=B_1-B_2。两者重量相差超过 11 克即 ∣X∣>1|X|>1。由于 XX 的分布关于 00 对称，可只求右侧尾部 P(X>1)P(X>1)，再乘以 22。

答题过程

展开

Let B1B_1 and B2B_2 be the weights of the two independently selected buns, and let

X=B1−B2.X=B_1-B_2.

Then

E(X)= 55−55=0,Var⁡(X)= 1.32+1.32=3.38.\begin{align*} E(X)=&\,55-55=0,\\ \operatorname{Var}(X)=&\,1.3^2+1.3^2=3.38. \end{align*}

Therefore,

X∼N(0,3.38).X\sim N(0,3.38).

By symmetry,

P(∣X∣>1)= 2P(X>1)= 2P(Z>13.38)= 2P(Z>0.5439…)= 0.58649…\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

0.586.\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 VV、YY。每袋都包含一个独立的空袋重量，因此比较 Y−VY-V 时，两个空袋重量的方差都要计入。先求 VV 与 YY 的均值和方差，再令 D=Y−VD=Y-V；所求事件就是 D>0D>0。

答题过程

展开

Let SBS_B and SRS_R be the independent weights of the empty bags used for the buns and the rolls respectively. Let

V=B1+B2+⋯+B10+SBV=B_1+B_2+\cdots+B_{10}+S_B

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

Y=R1+R2+⋯+R11+SRY=R_1+R_2+\cdots+R_{11}+S_R

be the total weight of a bag of rolls.

For the bag of buns,

E(V)= 10(55)+3=553,Var⁡(V)= 10(1.32)+0.22=16.94.\begin{align*} E(V)=&\,10(55)+3=553,\\ \operatorname{Var}(V)=&\,10(1.3^2)+0.2^2=16.94. \end{align*}

Hence

V∼N(553,16.94).V\sim N(553,16.94).

For the bag of rolls,

E(Y)= 11(51)+3=564,Var⁡(Y)= 11(1.22)+0.22=15.88.\begin{align*} E(Y)=&\,11(51)+3=564,\\ \operatorname{Var}(Y)=&\,11(1.2^2)+0.2^2=15.88. \end{align*}

Hence

Y∼N(564,15.88).Y\sim N(564,15.88).

Now let

D=Y−V.D=Y-V.

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

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

Therefore,

D∼N(11,32.82).D\sim N(11,32.82).

The required probability is

P(V<Y)= P(D>0)= P(Z>0−1132.82)= P(Z>−1.920…)= 0.9726…\begin{align*} P(V<Y)=&\,P(D>0)\\ =&\,P\bigg(Z>\frac{0-11}{\sqrt{32.82}}\bigg)\\ =&\,P(Z>-1.920\ldots)\\ =&\,0.9726\ldots \end{align*}

Thus, to 3 significant figures,

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