题目
The manager of a supermarket is investigating the number of complaints per day received from customers.
A random sample of 180 days is taken and the results are shown in the table below.
| Number of complaints per day | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 or more |
|---|---|---|---|---|---|---|---|---|
| Frequency | 12 | 28 | 37 | 38 | 29 | 17 | 19 | 0 |
(a) Calculate the mean and the variance of these data.
(b) Explain why the results in part (a) suggest that a Poisson distribution may be a suitable model for the number of complaints per day.
The manager uses a Poisson distribution with mean 3 to model the number of complaints per day.
(c) For a randomly selected day find, using the manager’s model, the probability that there are
(i) at least 3 complaints,
(ii) more than 4 complaints but less than 8 complaints.
A week consists of 7 consecutive days.
(d) Using the manager’s model and a suitable approximation, show that the probability that there are less than 19 complaints in a randomly selected week is 0.29 to 2 decimal places.
Show your working clearly.
(Solutions relying on calculator technology are not acceptable.)
A period of 13 weeks is selected at random.
(e) Find the probability that in this period there are exactly 5 weeks that have less than 19 complaints.
Show your working clearly.
(Total for Question 1 is 16 marks)
题目中文翻译
超市经理正在调查每天收到的客户投诉数量。
随机抽取 180 天作为样本，结果如下表所示。
| 每天投诉数 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 或以上 |
|---|---|---|---|---|---|---|---|---|
| 频数 | 12 | 28 | 37 | 38 | 29 | 17 | 19 | 0 |
(a) 计算这些数据的均值和方差。
(b) 解释为什么 (a) 中的结果表明泊松分布可能是每天投诉数的合适模型。
经理使用均值为 3 的泊松分布来建模每天的投诉数。
(c) 对于随机选择的一天，使用经理的模型求以下情况的概率
(i) 至少 3 次投诉，
(ii) 超过 4 次但少于 8 次投诉。
一周由 7 个连续天组成。
(d) 使用经理的模型和合适的近似方法，证明在随机选择的一周内投诉数少于 19 次的概率为 0.29（保留 2 位小数）。
清楚地展示你的计算过程。 （依赖计算器技术的解答不可接受。）
随机选择一个 13 周的时间段。
(e) 求在此期间恰好有 5 周的投诉数少于 19 次的概率。
清楚地展示你的计算过程。
（第 1 题共 16 分）