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S1 2021 June Q4

A Level / Edexcel / S1

IAL 2021 June Paper · Question 4

题目

Problem

Kris works in the mailroom of a large company and is responsible for all the letters sent by the company. The weights of letters sent by the company, WW grams, have a normal distribution with mean 165 g and standard deviation 35 g.

(a) Estimate the proportion of letters sent by the company that weigh less than 120 g.

(3)

Kris splits the letters to be sent into 3 categories: heavy, medium and light, with 13\frac13 of the letters in each category.

(b) Find the weight limits that determine medium letters.

(4)

A heavy letter is chosen at random.

(c) Find the probability that this letter weighs less than 200 g.

(3)

Kris chooses a random sample of 3 letters from those in the mailroom one day.

(d) Find the probability that there is one letter in each of the 3 categories.

(3)
题目中文翻译

Kris 在一家大公司的收发室工作，负责公司寄出的所有信件。公司寄出信件的重量 WW 克服从均值为 165 g、标准差为 35 g 的正态分布。

(a) 估计公司寄出的信件中重量小于 120 g 的比例。

Kris 将待寄信件分为重、中、轻三类，每类各占信件总数的 13\frac13。

(b) 求界定中等重量信件的重量界限。

随机选取一封重信件。

(c) 求该信件重量小于 200 g 的概率。

某天，Kris 从收发室的信件中随机抽取 3 封。

(d) 求这 3 封信件中三类各有一封的概率。

解答

(a)

解法一

思路

展开

先标准化，再查 Φ(z)\Phi(z)。因为 120120 在平均数左边，所以最后要取左尾概率。

答题过程

展开 P(W<120)=P(Z<120−16535)=P(Z<−1.2857…).P(W<120) =P\left(Z<\frac{120-165}{35}\right) =P(Z<-1.2857\ldots).

Using symmetry,

P(Z<−1.2857…)=1−P(Z<1.2857…).\begin{align*} P(Z<-1.2857\ldots)=1-P(Z<1.2857\ldots). \end{align*}

From tables,

P(Z<1.29)≈0.9015.\begin{align*} P(Z<1.29)\approx0.9015. \end{align*}

Therefore

P(W<120)≈1−0.9015=0.0985.\begin{align*} P(W<120)\approx1-0.9015=0.0985. \end{align*}

(b)

解法一

思路

展开

三类各占三分之一，所以 medium 是中间三分之一。边界对应标准正态的下三分之一点和上三分之一点，即 z≈±0.43z\approx\pm0.43。

答题过程

展开

The medium letters are between the lower one-third point and the upper one-third point.

Using the percentage points of the normal distribution,

z≈0.43.\begin{align*} z\approx0.43. \end{align*}

So the two limits are found from

x−16535=±0.43.\begin{align*} \frac{x-165}{35}=\pm0.43. \end{align*}

Hence

x=165±0.43(35).\begin{align*} x=165\pm0.43(35). \end{align*}

Therefore

x=165±15.05.\begin{align*} x=165\pm15.05. \end{align*}

The weight limits are approximately

150 g and 180 g.\begin{align*} 150\text{ g and }180\text{ g}. \end{align*}

(c)

解法一

思路

展开

已知选到的是 heavy letter，所以分母是 P(W>180)=13P(W>180)=\frac13。要小于 200，则分子是 180<W<200180<W<200。

答题过程

展开

Using the upper limit for medium letters from part (b), a heavy letter has W>180W>180.

Therefore

P(W<200∣W>180)=P(180<W<200)P(W>180).P(W<200\mid W>180) =\frac{P(180<W<200)}{P(W>180)}.

Now

200−16535=1.\begin{align*} \frac{200-165}{35}=1. \end{align*}

So

P(W<200)=P(Z<1)=0.8413.\begin{align*} P(W<200)=P(Z<1)=0.8413. \end{align*}

Also P(W≤180)≈23P(W\leq180)\approx\frac23, so

P(180<W<200)=0.8413−23.\begin{align*} P(180<W<200)=0.8413-\frac23. \end{align*}

Hence

P(W<200∣W>180)= 0.8413−2313= 0.524….\begin{align*} P(W<200\mid W>180) =&\,\frac{0.8413-\frac23}{\frac13}\\ =&\,0.524\ldots. \end{align*}

Therefore the probability is approximately

0.524.\begin{align*} 0.524. \end{align*}

解法二

思路

展开

也可以在“已知是重信件”的条件下求 W≥200W\geq200 的概率，再取补集。由于 W≥200W\geq200 已包含 W>180W>180，条件概率的分子可直接写成 P(W≥200)P(W\geq200)。

答题过程

展开

Using the complement rule,

P(W<200∣W>180)= 1−P(W≥200∣W>180).\begin{align*} P(W<200\mid W>180) =&\,1-P(W\geq200\mid W>180). \end{align*}

Since W≥200W\geq200 is a subset of W>180W>180,

P(W≥200∣W>180)= P(W≥200)P(W>180).\begin{align*} P(W\geq200\mid W>180) =&\,\frac{P(W\geq200)}{P(W>180)}. \end{align*}

Now

P(W≥200)= P(Z≥200−16535)= P(Z≥1)= 1−0.8413= 0.1587.\begin{align*} P(W\geq200) =&\,P\left(Z\geq\frac{200-165}{35}\right)\\ =&\,P(Z\geq1)\\ =&\,1-0.8413\\ =&\,0.1587. \end{align*}

Since P(W>180)=13P(W>180)=\frac13,

P(W<200∣W>180)= 1−0.15871/3= 0.5239≈ 0.524.\begin{align*} P(W<200\mid W>180) =&\,1-\frac{0.1587}{1/3}\\ =&\,0.5239\\ \approx&\,0.524. \end{align*}

(d)

解法一

思路

展开

每一类的概率都是 13\frac13。三封信要恰好一封属于每类，先写某一个顺序的概率，再乘以 3!3! 个排列。

答题过程

展开

For one particular order, for example heavy, medium, light, the probability is

13⋅13⋅13=127.\begin{align*} \frac13\cdot\frac13\cdot\frac13=\frac{1}{27}. \end{align*}

There are

3!=6\begin{align*} 3!=6 \end{align*}

possible orders.

Therefore

P(one in each category)=6⋅127=29.P(\text{one in each category}) =6\cdot\frac{1}{27} =\frac{2}{9}.