题目

Problem

(i) As part of a recruitment exercise candidates are required to complete three separate tasks. The times taken, $A$ , $B$ and $C$ , in minutes, for candidates to complete the three tasks are such that

$A \sim N(21, 2^2)$ , $B \sim N(32, 7^2)$ and $C \sim N(45, 9^2)$

The time taken by an individual candidate to complete each task is assumed to be independent of the time taken to complete each of the other tasks.

A candidate is selected at random.

(a) Find the probability that the candidate takes a total time of more than 90 minutes to complete all three tasks.

(4)

(b) Find $P(A > B)$

(4)

(ii) A simple random sample, $X_1, X_2, X_3, X_4$ , is taken from a normal population with mean $\mu$ and standard deviation $\sigma$

Given that

$\overline{X} = \frac{X_1 + X_2 + X_3 + X_4}{4}$

and that

$P(X_1 > \overline{X} + k\sigma) = 0.1$

where $k$ is a constant,

find the value of $k$ , giving your answer correct to 3 significant figures. You should show all stages of your working.

(7)

(Total for Question 7 is 15 marks)

(Total for Paper is 75 marks)

题目中文翻译

(i) 作为招聘活动的一部分，候选人需要完成三个独立任务。完成三个任务所用的时间分别为 A、B 和 C（单位：分钟），且

A ~ N(21, 2^2)，B ~ N(32, 7^2)，C ~ N(45, 9^2)

假设候选人完成每个任务所需时间彼此独立。

随机选取一名候选人。

(a) 求该候选人完成全部三个任务所用总时间超过 90 分钟的概率。

(b) 求 P(A > B)。

(ii) 从一个均值为 μ、标准差为 σ 的正态总体中抽取一个简单随机样本 X1, X2, X3, X4。

已知

X̄ = (X1 + X2 + X3 + X4) / 4

并且

P(X1 > X̄ + kσ) = 0.1

其中 k 为常数。

求 k 的值，答案保留 3 位有效数字。你应写出全部过程。

解答

(i)(a)

解法一

思路

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令总时间 $T=A+B+C$ 。独立正态变量之和仍服从正态分布，其均值相加、方差相加。求出 $T$ 的分布后，将 90 标准化并计算上尾概率。

答题过程

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Let

T=A+B+C.

Since $A$ , $B$ and $C$ are independent normal random variables,

\begin{align*} E(T) =&\,21+32+45\\ =&\,98, \end{align*}

and

\begin{align*} \operatorname{Var}(T) =&\,2^2+7^2+9^2\\ =&\,134. \end{align*}

Therefore,

T\sim\operatorname{N}(98,134).

Hence,

\begin{align*} P(T>90) =&\,P\left(Z>\frac{90-98}{\sqrt{134}}\right)\\ =&\,P(Z>-0.69109\ldots)\\ =&\,\boxed{0.755}\quad\text{(3 s.f.)}. \end{align*}

(i)(b)

解法一

思路

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把事件 $A>B$ 改写为 $A-B>0$ 。由于 $A$ 、 $B$ 独立，差 $A-B$ 仍为正态变量；其均值相减，但方差仍相加。确定差的分布后计算超过 0 的概率。

答题过程

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Let

D=A-B.

Since $A$ and $B$ are independent,

\begin{align*} E(D) =&\,21-32\\ =&\,-11, \end{align*}

and

\begin{align*} \operatorname{Var}(D) =&\,2^2+7^2\\ =&\,53. \end{align*}

Thus,

D\sim\operatorname{N}(-11,53).

Therefore,

\begin{align*} P(A>B) =&\,P(D>0)\\ =&\,P\left(Z>\frac{0-(-11)}{\sqrt{53}}\right)\\ =&\,P(Z>1.51097\ldots)\\ =&\,\boxed{0.0654}\quad\text{(3 s.f.)}. \end{align*}

(ii)

解法一

思路

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先把原事件改写为 $X_1-\overline X>k\sigma$ 。由于 $\overline X$ 中含有 $X_1$ ，不能错误地把 $X_1$ 与 $\overline X$ 当成独立变量；应先将差展开成 $X_1,X_2,X_3,X_4$ 的线性组合，再利用样本各项独立来求其均值和方差。最后用上尾概率 $0.1$ 对应的标准正态分位数 $1.2816$ 求 $k$ 。

答题过程

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Define

D=X_1-\overline{X}.

Then

\begin{align*} D =&\,X_1-\frac{X_1+X_2+X_3+X_4}{4}\\ =&\,\frac{3X_1-X_2-X_3-X_4}{4}. \end{align*}

Since $X_1,X_2,X_3,X_4$ are independent and each has mean $\mu$ and variance $\sigma^2$ ,

\begin{align*} E(D) =&\,\frac{3\mu-\mu-\mu-\mu}{4}\\ =&\,0, \end{align*}

and

\begin{align*} \operatorname{Var}(D) =&\,\frac{9\sigma^2+\sigma^2+\sigma^2+\sigma^2}{4^2}\\ =&\,\frac{12\sigma^2}{16}\\ =&\,0.75\sigma^2. \end{align*}

As $D$ is a linear combination of independent normal random variables,

D\sim\operatorname{N}(0,0.75\sigma^2).

The given probability becomes

P(D>k\sigma)=0.1.

Since the standard deviation of $D$ is $\sqrt{0.75}\,\sigma$ ,

P\left(Z>\frac{k}{\sqrt{0.75}}\right)=0.1.

The standard normal value with upper-tail probability $0.1$ is $1.2816$ . Therefore,

\begin{align*} \frac{k}{\sqrt{0.75}} =&\,1.2816\\ k =&\,\sqrt{0.75}(1.2816)\\ =&\,1.10990\ldots\\ =&\,\boxed{1.11}\quad\text{(3 s.f.)}. \end{align*}

IAL 2026 Jan S3 A Q7

题目

解答

(i)(a)

解法一

思路

答题过程

(i)(b)

解法一

思路

答题过程

(ii)

解法一

思路

答题过程