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IAL 2019 Oct S2 Q2

A Level / Edexcel / S2

IAL 2019 Oct Paper · Question 2

题目

Problem

During morning hours, employees arrive randomly at an office drinks dispenser at a rate of 2 every 10 minutes.

The number of employees arriving at the drinks dispenser is assumed to follow a Poisson distribution.

(a) Find the probability that fewer than 5 employees arrive at the drinks dispenser during a 10-minute period one morning.

(2)

During a 30-minute period one morning, the probability that nn employees arrive at the drinks dispenser is the same as the probability that n+1n+1 employees arrive at the drinks dispenser.

(b) Find the value of nn

(3)

During a 45-minute period one morning, the probability that between cc and 12, inclusive, employees arrive at the drinks dispenser is 0.8546

(c) Find the value of cc

(3)

(d) Find the probability that exactly 2 employees arrive at the drinks dispenser in exactly 4 of the 6 non-overlapping 10-minute intervals between 10 am and 11 am one morning.

(4)

(Total for Question 2 is 12 marks)

题目中文翻译

在上午时段，员工随机到达办公室的饮料机，平均每 10 分钟到达 2 人。

假设到达饮料机的员工人数服从泊松分布。

(a) 求某个上午的 10 分钟时段内，少于 5 名员工到达饮料机的概率。

某个上午的 30 分钟时段内，恰有 nn 名员工到达饮料机的概率，与恰有 n+1n+1 名员工到达的概率相同。

(b) 求 nn 的值。

某个上午的 45 分钟时段内，到达饮料机的员工人数在 cc 到 12 之间（含端点）的概率为 0.8546。

(c) 求 cc 的值。

(d) 某个上午 10 时至 11 时包含 6 个互不重叠的 10 分钟时段。求其中恰好 4 个时段各有恰好 2 名员工到达饮料机的概率。

（第 2 题共 12 分）

解答

(a)

解法一

思路

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10 分钟内的平均到达人数为 2，因此使用参数为 2 的泊松分布。“少于 5 人”等价于人数不超过 4。

答题过程

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Let XX be the number of arrivals in 10 minutes. Then

X∼Po⁡(2).X\sim\operatorname{Po}(2).

Therefore,

P(X<5)= P(X⩽4)= 0.947346…= 0.947(3 s.f.).\begin{align*} P(X<5) =&\,P(X\leqslant4) \\[4mm] =&\,0.947346\ldots \\[4mm] =&\,\boxed{0.947}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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30 分钟的泊松均值为 6。把 P(Y=n)P(Y=n) 与 P(Y=n+1)P(Y=n+1) 都写成泊松概率公式并令它们相等；约去公共因子后即可解出整数 nn。

答题过程

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Let YY be the number of arrivals in 30 minutes. Then

Y∼Po⁡(6).Y\sim\operatorname{Po}(6).

Given that P(Y=n)=P(Y=n+1)P(Y=n)=P(Y=n+1),

e−66nn!=e−66n+1(n+1)!.e^{-6}\frac{6^n}{n!} =e^{-6}\frac{6^{n+1}}{(n+1)!}.

Cancelling the common factors gives

1= 6n+1n+1= 6n= 5.\begin{align*} 1=&\,\frac6{n+1} \\[4mm] n+1=&\,6 \\[4mm] n=&\,\boxed{5}. \end{align*}

(c)

解法一

思路

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45 分钟的泊松均值为 9。把闭区间概率写成两个累积概率之差，由已知总概率反推出下端点之前的累积概率，再查泊松分布得到相应整数。

答题过程

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Let DD be the number of arrivals in 45 minutes. Then

D∼Po⁡(9).D\sim\operatorname{Po}(9).

Now,

P(c⩽D⩽12)= P(D⩽12)−P(D⩽c−1)0.8546= 0.8758−P(D⩽c−1).\begin{align*} P(c\leqslant D\leqslant12) =&\,P(D\leqslant12)-P(D\leqslant c-1) \\[4mm] 0.8546 =&\,0.8758-P(D\leqslant c-1). \end{align*}

Hence,

P(D⩽c−1)=0.0212.P(D\leqslant c-1)=0.0212.

For D∼Po⁡(9)D\sim\operatorname{Po}(9),

P(D⩽3)=0.0212(4 d.p.).P(D\leqslant3)=0.0212\quad\text{(4 d.p.)}.

Therefore, c−1=3c-1=3, so

c=4.\boxed{c=4}.

(d)

解法一

思路

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先求任一 10 分钟时段内恰有 2 人到达的概率。泊松过程在互不重叠时段内的到达人数相互独立，因此 6 个时段中满足该事件的时段数服从二项分布，再求恰好 4 个时段满足条件的概率。

答题过程

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For one 10-minute interval,

p=P(X=2)= e−2222!= 0.270671…\begin{align*} p=P(X=2) =&\,e^{-2}\frac{2^2}{2!} \\[4mm] =&\,0.270671\ldots \end{align*}

Let WW be the number of the 6 intervals in which exactly 2 employees arrive. Then

W∼B⁡(6,p).W\sim\operatorname{B}(6,p).

Therefore,

P(W=4)= (64)p4(1−p)2= 0.0428…= 0.043(3 s.f.).\begin{align*} P(W=4) =&\,\binom64p^4(1-p)^2 \\[4mm] =&\,0.0428\ldots \\[4mm] =&\,\boxed{0.043}\quad\text{(3 s.f.)}. \end{align*}