题目

Problem

The weights of bags of carrots, C kg, are such that C ~ N(1.2, 0.03^2) Three bags of carrots are selected at random.

(a) Calculate the probability that their total weight is more than 3.5 kg.

(4)

The weights of bags of potatoes, R kg, are such that R ~ N(2.3, 0.03^2) Two bags of potatoes are selected at random.

(b) Calculate the probability that the difference in their weights is more than 0.05 kg.

(5)

The weights of trays, $T$ kg, are such that $T\sim\operatorname{N}(2.5,\sqrt{0.1^2})$

The random variable G represents the total weight, in kg, of a single tray packed with 10 bags of potatoes where G and T are independent.

(7)

(Total for Question 6 is 16 marks)

题目中文翻译

胡萝卜袋的重量 C kg 服从分布 C ~ N(1.2, 0.03^2)。随机抽取 3 袋胡萝卜。

(a) 计算它们总重量大于 3.5 kg 的概率。

土豆袋的重量 R kg 服从分布 R ~ N(2.3, 0.03^2)。随机抽取 2 袋土豆。

(b) 计算它们重量差大于 0.05 kg 的概率。

托盘的重量 $T$ kg 服从分布 $T\sim\operatorname{N}(2.5,\sqrt{0.1^2})$ 。

随机变量 G 表示一个装有 10 袋土豆的托盘的总重量（单位 kg），且 G 与 T 相互独立。

解答

(a)

解法一

思路

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三袋胡萝卜的重量相互独立，因此总重量仍服从正态分布；均值相加，方差也相加。得到总重量分布后，将 3.5 标准化并求右尾概率。

答题过程

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Let

X=C_1+C_2+C_3.

Since the bag weights are independent,

\begin{align*} E(X)=&\,3(1.2)=3.6,\\ \operatorname{Var}(X) =&\,3(0.03^2)\\ =&\,0.0027. \end{align*}

Thus

X\sim\operatorname{N}(3.6,0.0027).

Therefore,

\begin{align*} P(X>3.5) =&\,P\left( Z>\frac{3.5-3.6}{\sqrt{0.0027}} \right)\\ =&\,P(Z>-1.9245\ldots)\\ =&\,\boxed{0.973}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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“重量差超过 0.05”表示差的绝对值超过 0.05。令 $Y=R_1-R_2$ ，其均值为 0，且分布关于 0 对称，所以先求一个尾部概率，再乘以 2。

答题过程

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Let

Y=R_1-R_2.

Since $R_1$ and $R_2$ are independent,

\begin{align*} E(Y)=&\,2.3-2.3=0,\\ \operatorname{Var}(Y) =&\,0.03^2+(-1)^2(0.03^2)\\ =&\,0.0018. \end{align*}

Therefore,

\begin{align*} P(Y>0.05) =&\,P\left( Z>\frac{0.05}{\sqrt{0.0018}} \right)\\ =&\,P(Z>1.1785\ldots)\\ =&\,0.1193\ldots. \end{align*}

By symmetry,

\begin{align*} P(|R_1-R_2|>0.05) =&\,2P(Y>0.05)\\ =&\,2(0.1193\ldots)\\ =&\,\boxed{0.239}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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先求装有 10 袋土豆的托盘总重量 $G$ 的均值和方差。题目给出 $G$ 与另一个托盘重量 $T$ 独立，因此令 $W=G-2T$ ，把所求事件化为 $W<20$ 。注意正态分布第二参数为方差，题中的 $\sqrt{0.1^2}=0.1$ 。

答题过程

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The total weight $G$ consists of one tray and 10 bags of potatoes. Hence

\begin{align*} E(G) =&\,2.5+10(2.3)\\ =&\,25.5, \end{align*}

and, since $\sqrt{0.1^2}=0.1$ ,

\begin{align*} \operatorname{Var}(G) =&\,0.1+10(0.03^2)\\ =&\,0.109. \end{align*}

Let

W=G-2T.

Since $G$ and $T$ are independent,

\begin{align*} E(W) =&\,25.5-2(2.5)\\ =&\,20.5, \end{align*}

and

\begin{align*} \operatorname{Var}(W) =&\,0.109+(-2)^2(0.1)\\ =&\,0.509. \end{align*}

Thus

W\sim\operatorname{N}(20.5,0.509).

Therefore,

\begin{align*} P(G<2T+20) =&\,P(W<20)\\ =&\,P\left( Z<\frac{20-20.5}{\sqrt{0.509}} \right)\\ =&\,P(Z<-0.7008\ldots)\\ =&\,\boxed{0.242}\quad\text{(3 s.f.)}. \end{align*}

IAL 2024 June S3 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