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S3 2025 Jun Q8

A Level / Edexcel / S3

IAL 2025 jun Paper · Question 8

题目

Problem

8.

The weights, SS kg, of bags of sand are such that S∼N⁡(18,0.032)S\sim\operatorname{N}(18,0.03^2).

Three bags of sand are selected at random.

(a) Using standardisation, calculate the probability that their total weight is more than 54.1 kg.

(4)

The weights, CC kg, of bags of cement are such that C∼N⁡(25,0.032)C\sim\operatorname{N}(25,0.03^2).

Two bags of cement are selected at random.

(b) Using standardisation, calculate the probability that their weights differ by more than 0.02 kg.

(5)

The weights, PP kg, of pallets are such that P∼N⁡(15,0.22)P\sim\operatorname{N}(15,0.2^2).

The random variable TT represents the total weight, in kg, of 28 bags of sand with 5 bags of cement and a single pallet of weight P1P_1.

(c) Using standardisation, calculate the probability that T<30P1+190T<30P_1+190.

(6)
(Total for Question 8 is 15 marks)
题目中文翻译

沙袋的重量 SS kg 满足 S∼N⁡(18,0.032)S\sim\operatorname{N}(18,0.03^2)。

随机选取 3 个沙袋。

(a) 用标准化方法求它们总重量大于 54.1 kg 的概率。

(4)

水泥袋的重量 CC kg 满足 C∼N⁡(25,0.032)C\sim\operatorname{N}(25,0.03^2)。

随机选取 2 个水泥袋。

(b) 用标准化方法求它们重量相差超过 0.02 kg 的概率。

(5)

托盘的重量 PP kg 满足 P∼N⁡(15,0.22)P\sim\operatorname{N}(15,0.2^2)。

随机变量 TT 表示 28 个沙袋、5 个水泥袋和 1 个重量为 P1P_1 的托盘的总重量，单位为 kg。

(c) 用标准化方法求 T<30P1+190T<30P_1+190 的概率。

(6)

解答

(a)

解法一

思路

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三个随机沙袋重量相互独立，其总重量仍服从正态分布。总均值是三个均值之和，总方差是三个方差之和；确定新分布后把 54.1 标准化。

答题过程

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Let

X=S1+S2+S3.X=S_1+S_2+S_3.

Since the bag weights are independent,

E(X)= 3(18)=54,Var⁡(X)= 3(0.032)= 0.0027.\begin{align*} E(X)=&\,3(18)=54,\\ \operatorname{Var}(X) =&\,3(0.03^2)\\ =&\,0.0027. \end{align*}

Thus

X∼N⁡(54,0.0027).X\sim\operatorname{N}(54,0.0027).

Therefore,

P(X>54.1)= P(Z>54.1−540.0027)= P(Z>1.9245…)= 0.0271(3 s.f.).\begin{align*} P(X>54.1) =&\,P\big( Z>\frac{54.1-54}{\sqrt{0.0027}} \big)\\ =&\,P(Z>1.9245\ldots)\\ =&\,\boxed{0.0271}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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“两袋重量相差超过 0.02”是绝对值事件。令 Y=C1−C2Y=C_1-C_2，其均值为 0；正态分布关于 0 对称，因此先算 P(Y>0.02)P(Y>0.02)，再乘以 2。

答题过程

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Let

Y=C1−C2.Y=C_1-C_2.

Since C1C_1 and C2C_2 are independent,

E(Y)= 25−25=0,Var⁡(Y)= 0.032+(−1)2(0.032)= 0.0018.\begin{align*} E(Y)=&\,25-25=0,\\ \operatorname{Var}(Y) =&\,0.03^2+(-1)^2(0.03^2)\\ =&\,0.0018. \end{align*}

Therefore,

P(Y>0.02)= P(Z>0.020.0018)= P(Z>0.4714…)= 0.318675….\begin{align*} P(Y>0.02) =&\,P\big( Z>\frac{0.02}{\sqrt{0.0018}} \big)\\ =&\,P(Z>0.4714\ldots)\\ =&\,0.318675\ldots. \end{align*}

By symmetry,

P(∣C1−C2∣>0.02)= 2P(Y>0.02)= 2(0.318675…)= 0.637(3 s.f.).\begin{align*} P(|C_1-C_2|>0.02) =&\,2P(Y>0.02)\\ =&\,2(0.318675\ldots)\\ =&\,\boxed{0.637}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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先把含 P1P_1 的项移到同一边。由于 TT 本身已经含一个 P1P_1，所以 T−30P1T-30P_1 中托盘项的系数是 −29-29，不是 −30-30。随后求这个线性组合的均值、方差并标准化。

答题过程

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Write

T=P1+∑i=15Ci+∑i=128Si.T=P_1+\sum_{i=1}^{5}C_i+\sum_{i=1}^{28}S_i.

The required event is

T−30P1<190.T-30P_1<190.

Let

R= T−30P1= ∑i=15Ci+∑i=128Si−29P1.\begin{align*} R =&\,T-30P_1\\ =&\,\sum_{i=1}^{5}C_i +\sum_{i=1}^{28}S_i-29P_1. \end{align*}

Using independence,

E(R)= 5(25)+28(18)−29(15)= 194,\begin{align*} E(R) =&\,5(25)+28(18)-29(15)\\ =&\,194, \end{align*}

and

Var⁡(R)= 5(0.032)+28(0.032)+(−29)2(0.22)= 33.6697.\begin{align*} \operatorname{Var}(R) =&\,5(0.03^2)+28(0.03^2)\\ &\qquad{}+(-29)^2(0.2^2)\\ =&\,33.6697. \end{align*}

Hence

R∼N⁡(194,33.6697).R\sim\operatorname{N}(194,33.6697).

Therefore,

P(T<30P1+190)= P(R<190)= P(Z<190−19433.6697)= P(Z<−0.6893…)= 0.245(3 s.f.).\begin{align*} P(T<30P_1+190) =&\,P(R<190)\\ =&\,P\big( Z<\frac{190-194}{\sqrt{33.6697}} \big)\\ =&\,P(Z<-0.6893\ldots)\\ =&\,\boxed{0.245}\quad\text{(3 s.f.)}. \end{align*}