题目

Problem

(a) Explain what you understand by the sampling distribution of a statistic.

(1)

At Sam’s cafe a standard breakfast consists of 6 breakfast items. Customers can then choose to upgrade to a medium breakfast by adding 1 extra breakfast item or they can upgrade to a large breakfast by adding 2 extra breakfast items. Standard, medium and large breakfasts are sold in the ratio 6 : 3 : 2 respectively. A random sample of 2 customers is taken from customers who have bought a breakfast from Sam’s cafe on a particular day.

(b) Find the sampling distribution for the total number, $T$ , of breakfast items bought by these 2 customers. Show your working clearly.

(7)

(2)

(Total for Question 6 is 10 marks)

题目中文翻译

(a) 解释你对统计量抽样分布的理解。

Sam 的咖啡馆里，一份标准早餐包含 6 种早餐食物。顾客可以升级为中份早餐，多加 1 种食物，或者升级为大份早餐，多加 2 种食物。标准、中份和大份早餐的销售比例分别为 6 : 3 : 2。从某一天在 Sam 咖啡馆购买早餐的顾客中随机抽取 2 人。

(b) 求这 2 位顾客所购买早餐食物总数 $T$ 的抽样分布。清楚展示你的计算过程。

（第 6 题共 10 分）

解答

(a)

解法一

思路

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抽样分布不仅要列出统计量所有可能的取值，还要给出各取值在随机抽样下对应的概率。回答中必须同时体现“统计量的取值”和“相关概率”。

答题过程

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The sampling distribution of a statistic is the probability distribution of that statistic under random sampling; it gives all possible values of the statistic and their associated probabilities.

(b)

解法一

思路

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一位顾客购买的食物数可能为 $6,7,8$ ，由销售比例 $6:3:2$ 可得对应概率。两位顾客是随机抽取的，因此列出两个有序结果并将各自概率相乘；总数相同的结果再合并。

答题过程

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For one customer, the number of breakfast items has probabilities

P(6)=\frac{6}{11},\qquad P(7)=\frac{3}{11},\qquad P(8)=\frac{2}{11}.

The possible ordered pairs are

\begin{matrix} (6,6)&(6,7)&(6,8)\\ (7,6)&(7,7)&(7,8)\\ (8,6)&(8,7)&(8,8). \end{matrix}

Therefore, the possible values of $T$ are $12,13,14,15,16$ . Their probabilities are

P(T=12)=\left(\frac{6}{11}\right)^2 =\frac{36}{121},

\begin{align*} P(T=13)=&\,2\left(\frac{6}{11}\right) \left(\frac{3}{11}\right)\\[2mm] =&\,\frac{36}{121}, \end{align*}

\begin{align*} P(T=14)=&\,2\left(\frac{6}{11}\right) \left(\frac{2}{11}\right)\\[2mm] +&\,\left(\frac{3}{11}\right)^2\\[2mm] =&\,\frac{33}{121}, \end{align*}

\begin{align*} P(T=15)=&\,2\left(\frac{3}{11}\right) \left(\frac{2}{11}\right)\\[2mm] =&\,\frac{12}{121}, \end{align*}

and

P(T=16)=\left(\frac{2}{11}\right)^2 =\frac{4}{121}.

Hence the sampling distribution is

$t$	$12$	$13$	$14$	$15$	$16$
$P(T=t)$	$\frac{36}{121}$	$\frac{36}{121}$	$\frac{33}{121}$	$\frac{12}{121}$	$\frac{4}{121}$

The probabilities sum to

\frac{36+36+33+12+4}{121}=1,

as required.

(c)

解法一

思路

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按照 (b) 得到的抽样分布，使用离散型随机变量的期望公式 $E(T)=\sum tP(T=t)$ ，逐项相乘后相加。

答题过程

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Using the sampling distribution from part (b),

\begin{align*} E(T)=&\,12\left(\frac{36}{121}\right) +13\left(\frac{36}{121}\right)\\[2mm] &\,\hspace{2pt}+14\left(\frac{33}{121}\right) +15\left(\frac{12}{121}\right)\\[2mm] &\,\hspace{4pt}+16\left(\frac{4}{121}\right)\\[2mm] =&\,\frac{1606}{121}\\[2mm] =&\,\frac{146}{11}\\[2mm] =&\,13.2727\ldots \end{align*}

Therefore,

\boxed{E(T)=\frac{146}{11}\approx 13.3}.

解法二

思路

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也可以先求一位顾客购买食物数的期望，再利用期望的可加性。两位顾客的总数期望等于单人期望的两倍；这种方法不需要先列出 $T$ 的完整分布。

答题过程

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Let $Y$ be the number of breakfast items bought by one customer. Then

\begin{align*} E(Y)=&\,6\left(\frac{6}{11}\right) +7\left(\frac{3}{11}\right)\\[2mm] &\,\hspace{2pt}+8\left(\frac{2}{11}\right)\\[2mm] =&\,\frac{73}{11}. \end{align*}

For two customers, linearity of expectation gives

\begin{align*} E(T)=&\,2E(Y)\\[2mm] =&\,2\left(\frac{73}{11}\right)\\[2mm] =&\,\boxed{\frac{146}{11}\approx 13.3}. \end{align*}

IAL 2022 Jan S2 Q6

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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解法二

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