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 $L$ cm, where $L \sim N\left(\mu, \, 0.2^2\right)$

The value of $\mu$ 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 $\mu$ 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 rods。这些 rods 从制造商处购买，制造商使用机器切割长度为 $L$ cm 的 rods，其中 $L \sim N\left(\mu, \, 0.2^2\right)$

$\mu$ 的值使得从商店供应的 rods 中随机选取一根，其长度小于 200 cm 的概率仅为 5%。

(a) 求 $\mu$ 的值，保留一位小数。

一位顾客随机购买了 8 根这样的 rods。

(b) 求其中至少有 3 根 rods 长度小于 200 cm 的概率。

另一位顾客随机购买了 60 根这样的 rods。

（第 1 题共 9 分）

解答

(a)

解法一

思路

展开

长度小于 $200$ cm 的概率是 $0.05$ ，对应标准正态分布的下侧 $5\%$ 分位数 $-1.6449$ 。将 $L$ 标准化后建立方程，求出总体均值 $\mu$ 。

答题过程

展开

Since $P(L<200)=0.05$ ,

P\left(Z<\frac{200-\mu}{0.2}\right)=0.05.

Using the lower $5\%$ standard normal quantile,

\begin{align*} \frac{200-\mu}{0.2}=&\,-1.6449\\ 200-\mu=&\,-0.32898\\ \mu=&\,200.32898. \end{align*}

Therefore,

\boxed{\mu=200.3\text{ cm}}\quad\text{(1 d.p.)}.

(b)

解法一

思路

展开

每根随机抽取的杆长度小于 $200$ cm 的概率为 $0.05$ 。令 $X$ 表示 8 根中长度小于 $200$ cm 的根数，则 $X$ 服从二项分布。用补事件把“至少 3 根”改写为 $1-P(X\leq2)$ 。

答题过程

展开

Let $X$ be the number of rods, out of 8, whose lengths are less than $200$ cm. Then

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

Hence,

\begin{align*} P(X\geq3) =&\,1-P(X\leq2)\\ =&\,1-0.994211\ldots\\ =&\,0.005788\ldots\\ =&\,\boxed{0.0058}\quad\text{(approximately)}. \end{align*}

(c)

解法一

思路

展开

60 根中长度小于 $200$ cm 的根数原本服从 $B(60,0.05)$ 。由于样本量较大而成功概率较小，可用参数 $np=3$ 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 $Y$ be the number of rods, out of 60, whose lengths are less than $200$ cm. Since

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

use the approximation

Y\mathrel{\dot\sim}\operatorname{Po}(60\times0.05) =\operatorname{Po}(3).

Therefore,

\begin{align*} P(Y>5) \approx&\,1-P(Y\leq5)\\ =&\,1-0.916082\ldots\\ =&\,0.083917\ldots\\ =&\,\boxed{0.0839}. \end{align*}

IAL 2026 Jan S2 A Q1

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