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IAL 2021 Jan Q3

A Level / Edexcel / S1

IAL 2021 Jan Paper · Question 3

题目

Problem

The weights of packages that arrive at a factory are normally distributed with a mean of 18 kg and a standard deviation of 5.4 kg

(a) Find the probability that a randomly selected package weighs less than 10 kg

(3)

The heaviest 15% of packages are moved around the factory by Jemima using a forklift truck.

(b) Find the weight, in kg, of the lightest of these packages that Jemima will move.

(3)

One of the packages not moved by Jemima is selected at random.

(c) Find the probability that it weighs more than 18 kg

(4)

A delivery of 4 packages is made to the factory. The weights of the packages are independent.

(d) Find the probability that exactly 2 of them will be moved by Jemima.

(3)

解答

(a)

解法一

思路

展开

先标准化 1010，再查左尾概率。

答题过程

展开

Let WW be the weight of a package.

W∼N(18,5.42).\begin{align*} W\sim N(18,5.4^2). \end{align*}

Then

P(W<10)=P(Z<10−185.4)=P(Z<−1.481…).P(W<10)=P\left(Z<\frac{10-18}{5.4}\right) =P(Z<-1.481\ldots).

Using symmetry,

P(W<10)=1−P(Z<1.481…)≈0.0694.\begin{align*} P(W<10)=1-P(Z<1.481\ldots)\approx0.0694. \end{align*}

(b)

解法一

思路

展开

最重的 15% 是右尾 0.150.15，对应标准正态 z=1.0364z=1.0364。

答题过程

展开

Let jj be the required weight. Since P(W>j)=0.15P(W>j)=0.15,

j−185.4=1.0364.\begin{align*} \frac{j-18}{5.4}=1.0364. \end{align*}

Therefore

j=18+1.0364(5.4)=23.596….\begin{align*} j=18+1.0364(5.4)=23.596\ldots. \end{align*}

So the lightest package Jemima will move weighs

23.6 kg.\begin{align*} 23.6\text{ kg}. \end{align*}

(c)

解法一

思路

展开

没有被 Jemima 移动表示 W<23.6W<23.6，概率为 0.85。要求其中 W>18W>18 的条件概率。

答题过程

展开 P(W>18∣W<23.6)=P(18<W<23.6)P(W<23.6).P(W>18\mid W<23.6) =\frac{P(18<W<23.6)}{P(W<23.6)}.

Now

P(W<23.6)=0.85\begin{align*} P(W<23.6)=0.85 \end{align*}

and

P(18<W<23.6)=0.85−0.50=0.35.\begin{align*} P(18<W<23.6)=0.85-0.50=0.35. \end{align*}

Hence

P(W>18∣W<23.6)=0.350.85=717.\begin{align*} P(W>18\mid W<23.6)=\frac{0.35}{0.85}=\frac{7}{17}. \end{align*}

解法二

思路

展开

条件概率补集法。 已知没有被 Jemima 移动的包裹满足 W<23.6W < 23.6（从第一问已知其概率为 0.850.85）。我们要求 P(W>18∣W<23.6)P(W > 18 \mid W < 23.6)，可以先求其对立事件（补集法则）：

P(W⩽18∣W<23.6)=P(W⩽18∩W<23.6)P(W<23.6)\begin{align*} P(W \leqslant 18 \mid W < 23.6) = \frac{P(W \leqslant 18 \cap W < 23.6)}{P(W < 23.6)} \end{align*}

由于均值 18<23.618 < 23.6，事件 W⩽18W \leqslant 18 完全包含在 W<23.6W < 23.6 之中。因此，交集直接简化为 P(W⩽18)P(W \leqslant 18)。 又因为 1818 正好是正态分布的均值 μ\mu，所以 P(W⩽18)=0.5P(W \leqslant 18) = 0.5。 代入后可得：

P(W⩽18∣W<23.6)=0.50.85=1017\begin{align*} P(W \leqslant 18 \mid W < 23.6) = \frac{0.5}{0.85} = \frac{10}{17} \end{align*}

最后，原条件概率即为 1−1017=7171 - \frac{10}{17} = \frac{7}{17}。这种求补集的方法将分子直接简化为对对称中心（均值）的单尾概率，在代数上比用区间概率相减要更优雅。

答题过程

展开

Use the complement rule for conditional probability:

P(W>18∣W<23.6)=  1−P(W⩽18∣W<23.6).\begin{align*} P(W > 18 \mid W < 23.6) =&\,\, 1 - P(W \leqslant 18 \mid W < 23.6). \end{align*}

Since 18<23.618 < 23.6, the event W⩽18W \leqslant 18 is a subset of W<23.6W < 23.6. Therefore:

P(W⩽18∩W<23.6)=P(W⩽18).\begin{align*} P(W \leqslant 18 \cap W < 23.6) = P(W \leqslant 18). \end{align*}

By definition of conditional probability:

P(W⩽18∣W<23.6)=  P(W⩽18)P(W<23.6).\begin{align*} P(W \leqslant 18 \mid W < 23.6) =&\,\, \frac{P(W \leqslant 18)}{P(W < 23.6)}. \end{align*}

Since 1818 is the mean of the normal distribution, we have P(W⩽18)=0.5P(W \leqslant 18) = 0.5. From part (b), the probability that a package is not moved by Jemima is:

P(W<23.6)=1−0.15=0.85.\begin{align*} P(W < 23.6) = 1 - 0.15 = 0.85. \end{align*}

Thus:

P(W⩽18∣W<23.6)=  0.50.85=  1017.\begin{align*} P(W \leqslant 18 \mid W < 23.6) =&\,\, \frac{0.5}{0.85}\\[3mm] =&\,\, \frac{10}{17}. \end{align*}

Hence:

P(W>18∣W<23.6)=  1−1017=  717.\begin{align*} P(W > 18 \mid W < 23.6) =&\,\, 1 - \frac{10}{17}\\[3mm] =&\,\, \frac{7}{17}. \end{align*}

(d)

解法一

思路

展开

每个 package 被移动的概率为 0.15，四个中恰好两个被移动，用排列数或组合数乘上对应概率。

答题过程

展开 P(exactly 2 moved)=(42)(0.15)2(0.85)2.P(\text{exactly 2 moved}) =\binom{4}{2}(0.15)^2(0.85)^2.

Therefore

P(exactly 2 moved)=6(0.15)2(0.85)2=0.0975375.P(\text{exactly 2 moved}) =6(0.15)^2(0.85)^2 =0.0975375.

So the probability is

0.0975to 3 significant figures.\begin{align*} 0.0975\quad\text{to 3 significant figures}. \end{align*}