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IAL 2024 Jun S2 Q1

A Level / Edexcel / S2

IAL 2024 June Paper · Question 1

题目

Problem

A garage sells tyres. The number of customers arriving at the garage to buy tyres in a 10-minute period is modelled by a Poisson distribution with mean 2

(a) Find the probability that

(i) fewer than 4 customers arrive to buy tyres in the next 10 minutes,

(ii) more than 5 customers arrive to buy tyres in the next 10 minutes.

(3)

The manager randomly selects 20 non-overlapping, 30-minute periods.

(b) Find the probability that there are between 4 and 7 (inclusive) customers arriving to buy tyres in exactly 15 of these 30-minute periods.

(4)

The manager believes that placing an advert in the local paper will lead to a significant increase in the number of customers arriving at the garage.

A week after the advert is placed, the manager randomly selects a 25-minute period and finds that 10 customers arrive at the garage to buy tyres.

(c) Test, at the 5% level of significance, whether or not there is evidence to support the manager’s belief. State your hypotheses clearly.

(5)

(d) Explain why the Poisson distribution is unlikely to be valid for the number of tyres sold during a 10-minute period.

(1)

(Total for Question 1 is 13 marks)

题目中文翻译

一家车库销售轮胎。10 分钟内到达车库购买轮胎的顾客数量用均值为 2 的泊松分布建模。

(a) 求以下情况的概率

(i) 在接下来 10 分钟内少于 4 名顾客到达购买轮胎，

(ii) 在接下来 10 分钟内超过 5 名顾客到达购买轮胎。

经理随机选择了 20 个不重叠的 30 分钟时间段。

(b) 求在这些 30 分钟时间段中恰好有 15 个时间段内到达购买轮胎的顾客数在 4 到 7（含）之间的概率。

经理认为在当地报纸上刊登广告将导致到达车库的顾客数量显著增加。

广告刊登一周后，经理随机选择了一个 25 分钟时间段，发现有 10 名顾客到达车库购买轮胎。

(c) 在 5% 的显著性水平下，检验是否有证据支持经理的信念。清楚地陈述你的假设。

(d) 解释为什么泊松分布不太可能适用于 10 分钟内销售的轮胎数量。

（第 1 题共 13 分）

解答

(a)(i)

解法一

思路

展开

10 分钟内的顾客数服从均值为 2 的泊松分布。“少于 4”即顾客数不超过 3，直接求累积概率。

答题过程

展开

Let MM be the number of customers arriving in 10 minutes. Then

M∼Po⁡(2).M\sim\operatorname{Po}(2).

Therefore,

P(M<4)=P(M⩽3)=0.8571…=0.857(3 s.f.).P(M<4)=P(M\leqslant3) =0.8571\ldots =\boxed{0.857}\quad\text{(3 s.f.)}.

(a)(ii)

解法一

思路

展开

“超过 5”即至少 6。利用补事件，从 1 减去顾客数不超过 5 的累积概率。

答题过程

展开 P(M>5)= 1−P(M⩽5)= 0.016563…= 0.0166(3 s.f.).\begin{align*} P(M>5)=&\,1-P(M\leqslant5)\\ =&\,0.016563\ldots\\ =&\,\boxed{0.0166}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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先按时间比例把 30 分钟内的泊松均值换成 6，求单个时段内顾客数在 4 到 7 之间的概率 pp。20 个互不重叠的时段可视为独立试验，因此满足该条件的时段数服从 B⁡(20,p)\operatorname{B}(20,p)，最后求恰好 15 个时段成功的概率。

答题过程

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Let QQ be the number of customers arriving in a 30-minute period. Then

Q∼Po⁡(6).Q\sim\operatorname{Po}(6).

For one period,

p=P(4⩽Q⩽7)= P(Q⩽7)−P(Q⩽3)= 0.5928…\begin{align*} p=P(4\leqslant Q\leqslant7) =&\,P(Q\leqslant7)-P(Q\leqslant3)\\ =&\,0.5928\ldots \end{align*}

Let XX be the number of the 20 periods satisfying this condition. Since the periods do not overlap,

X∼B⁡(20,p).X\sim\operatorname{B}(20,p).

Hence

P(X=15)= (2015)p15(1−p)5= 0.06807…= 0.0681(3 s.f.).\begin{align*} P(X=15)=&\,\binom{20}{15}p^{15}(1-p)^5\\ =&\,0.06807\ldots\\ =&\,\boxed{0.0681}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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经理认为到达率增加，因此使用右尾检验。原到达率是每 10 分钟 2 人，所以在原假设下，25 分钟的均值为 5。求观测到至少 10 人的右尾概率，并与 5%5\% 比较，最后用题目语境下结论。

答题过程

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Let λ\lambda be the mean number of customers arriving in a 10-minute period.

H0:λ=2,H1:λ>2.H_0:\lambda=2, \qquad H_1:\lambda>2.

Under H0H_0, let RR be the number of customers arriving in 25 minutes. Then

R∼Po⁡(2×2510)=Po⁡(5).R\sim\operatorname{Po}\left(2\times\frac{25}{10}\right) =\operatorname{Po}(5).

The pp-value is

P(R⩾10)= 1−P(R⩽9)= 0.0318…\begin{align*} P(R\geqslant10)=&\,1-P(R\leqslant9)\\ =&\,0.0318\ldots \end{align*}

Since 0.0318<0.050.0318<0.05, reject H0H_0. There is sufficient evidence, at the 5%5\% significance level, to support the manager’s belief that the rate of customers arriving at the garage has increased.

(d)

解法一

思路

展开

泊松模型要求事件单个发生且彼此独立，但顾客购买轮胎时常会一次购买一对或一整组，因此“售出一条轮胎”这类事件往往成批出现，不满足独立发生的假设。

答题过程

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Tyres are often bought in pairs or sets, so individual tyre sales are not independent and are unlikely to occur singly. Therefore, a Poisson model is unlikely to be valid for the number of tyres sold.