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IAL 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)

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{aligned} P(W<200\mid W>180) =&\,\frac{0.8413-\frac23}{\frac13}\\ =&\,0.524\ldots. \end{aligned}

Therefore the probability is approximately

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

解法二

思路

展开

条件概率补集法。 已知选到的是 heavy letter（即 W>180W > 180）。我们要求 P(W<200∣W>180)P(W < 200 \mid W > 180)，可以先求其对立事件（补集法则）：

P(W⩾200∣W>180)=P(W⩾200∩W>180)P(W>180)\begin{align*} P(W \geqslant 200 \mid W > 180) = \frac{P(W \geqslant 200 \cap W > 180)}{P(W > 180)} \end{align*}

由于 200>180200 > 180，事件 W⩾200W \geqslant 200 完全包含在 W>180W > 180 之中。因此，交集部分直接简化为单个变量概率：

P(W⩾200∩W>180)=P(W⩾200)\begin{align*} P(W \geqslant 200 \cap W > 180) = P(W \geqslant 200) \end{align*}

由此可得：

P(W⩾200∣W>180)=P(W⩾200)P(W>180)=1−P(W<200)1/3=3×P(W⩾200)\begin{align*} P(W \geqslant 200 \mid W > 180) = \frac{P(W \geqslant 200)}{P(W > 180)} = \frac{1 - P(W < 200)}{1/3} = 3 \times P(W \geqslant 200) \end{align*}

通过求补集，分子部分直接化简为单尾概率，省去了复杂的区间概率差 P(180<W<200)P(180 < W < 200) 的计算，在代数上极具优雅性，且避免了分式中循环小数 2/32/3 的截断误差。

答题过程

展开

Use the complement rule for conditional probability:

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

Since 200>180200 > 180, the event W⩾200W \geqslant 200 is a subset of W>180W > 180. Therefore:

P(W⩾200∩W>180)=P(W⩾200).\begin{align*} P(W \geqslant 200 \cap W > 180) = P(W \geqslant 200). \end{align*}

By definition of conditional probability:

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

Standardise 200200:

P(W⩾200)=  P(Z⩾200−16535)=  P(Z⩾1)=  1−0.8413=  0.1587.\begin{align*} P(W \geqslant 200) =&\,\, P\left(Z \geqslant \frac{200 - 165}{35}\right)\\[3mm] =&\,\, P(Z \geqslant 1)\\[3mm] =&\,\, 1 - 0.8413\\[3mm] =&\,\, 0.1587. \end{align*}

Since P(W>180)=13P(W > 180) = \frac{1}{3}, we have:

P(W⩾200∣W>180)=  0.15871/3=  3×0.1587=  0.4761.\begin{align*} P(W \geqslant 200 \mid W > 180) =&\,\, \frac{0.1587}{1/3}\\[3mm] =&\,\, 3 \times 0.1587\\[3mm] =&\,\, 0.4761. \end{align*}

Thus:

P(W<200∣W>180)=  1−0.4761=  0.5239.\begin{align*} P(W < 200 \mid W > 180) =&\,\, 1 - 0.4761\\[3mm] =&\,\, 0.5239. \end{align*}

Therefore the probability is approximately

0.524.\begin{align*} 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}.