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S2 2026 Jan A Q1

A Level / Edexcel / S2

IAL 2026 Jan A Paper · Question 1

题目

Problem

A shop sells rods of nominal length 200 cm. The rods are bought from a manufacturer who uses a machine to cut rods of length LL cm, where L∼N(μ, 0.22)L \sim N\left(\mu, \, 0.2^2\right)

The value of μ\mu is such that there is only a 5% chance that the rod, selected at random from those supplied to the shop, will have a length less than 200 cm.

(a) Find the value of μ\mu to one decimal place.

(3)

A customer buys a random sample of 8 of these rods.

(b) Find the probability that at least 3 of these rods will have a length less than 200 cm.

(3)

Another customer buys a random sample of 60 of these rods.

(c) Using a suitable approximation, find the probability that more than 5 of these rods will have length less than 200 cm.

(3)

(Total for Question 1 is 9 marks)

题目中文翻译

一家商店出售标称长度为 200 cm 的杆。这些杆从制造商处购买,制造商使用机器切割长度为 LL cm 的杆,其中 L∼N(μ, 0.22)L \sim N\big(\mu, \, 0.2^2\big)。

μ\mu 的值使得从供应给商店的杆中随机选取一根,其长度小于 200 cm 的概率仅为 5%。

(a) 求 μ\mu 的值,保留一位小数。

(3)

一位顾客随机购买了 8 根这样的杆。

(b) 求其中至少有 3 根杆长度小于 200 cm 的概率。

(3)

另一位顾客随机购买了 60 根这样的杆。

(c) 使用适当的近似方法,求其中超过 5 根杆长度小于 200 cm 的概率。

(3)

(第 1 题共 9 分)

解答

(a)

解法一

思路

展开

长度小于 200200 cm 的概率是 0.050.05,对应标准正态分布的下侧 5%5\% 分位数 −1.6449-1.6449。将 LL 标准化后建立方程,求出总体均值 μ\mu。

答题过程

展开

Let Z∼N⁡(0,1)Z\sim\operatorname{N}(0,1). Since P(L<200)=0.05P(L<200)=0.05,

P(Z<200−μ0.2)=0.05.P\Big(Z<\frac{200-\mu}{0.2}\Big)=0.05.

Using −1.6449-1.6449 as an approximation to the lower 5%5\% standard normal quantile,

200−μ0.2≈ −1.6449200−μ≈ −0.32898μ≈ 200.32898.\begin{align*} \frac{200-\mu}{0.2}\approx&\,-1.6449\\ 200-\mu\approx&\,-0.32898\\ \mu\approx&\,200.32898. \end{align*}

Therefore,

μ≈200.3 cm.\boxed{\mu\approx200.3\text{ cm}}.

The answer is given to one decimal place.

(b)

解法一

思路

展开

每根随机抽取的杆长度小于 200200 cm 的概率为 0.050.05。令 XX 表示 8 根中长度小于 200200 cm 的根数,则 XX 服从二项分布。用补事件把“至少 3 根”改写为 1−P(X≤2)1-P(X\leq2)。

答题过程

展开

Let XX be the number of rods, out of 8, whose lengths are less than 200200 cm. Then

X∼B⁡(8,0.05).X\sim\operatorname{B}(8,0.05).

Hence,

 P(X≥3)= 1−P(X≤2)= 1−0.994211…= 0.005788…≈ 0.0058.\begin{align*} &\,P(X\geq3)\\[4mm] =&\,1-P(X\leq2)\\ =&\,1-0.994211\ldots\\ =&\,0.005788\ldots\\ \approx&\,\boxed{0.0058}. \end{align*}

(c)

解法一

思路

展开

60 根中长度小于 200200 cm 的根数原本服从 B(60,0.05)B(60,0.05)。由于样本量较大而成功概率较小,可用参数 np=3np=3 的泊松分布近似;“超过 5 根”就是至少 6 根,使用补事件计算。

答题过程

展开

Let YY be the number of rods, out of 60, whose lengths are less than 200200 cm. Since

Y∼B⁡(60,0.05),Y\sim\operatorname{B}(60,0.05),

use the approximation

Y∼˙Po⁡(3),Y\mathrel{\dot\sim}\operatorname{Po}(3),

where 60×0.05=360\times0.05=3.

Therefore,

 P(Y>5)≈ 1−P(Y≤5)≈ 1−0.916082…≈ 0.083917…≈ 0.0839.\begin{align*} &\,P(Y>5)\\[4mm] \approx&\,1-P(Y\leq5)\\ \approx&\,1-0.916082\ldots\\ \approx&\,0.083917\ldots\\ \approx&\,\boxed{0.0839}. \end{align*}