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