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。

(1)

Sam 的咖啡馆里，一份标准早餐包含 6 件早餐食物。顾客可以升级为中份早餐，多加 1 件食物，或者升级为大份早餐，多加 2 件食物。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。

(7)

(2)

（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

\begin{align*} P(6)=&\,\frac6{11},\\[2mm] P(7)=&\,\frac3{11},\\[2mm] P(8)=&\,\frac2{11}. \end{align*}

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

\begin{align*} P(T=12)=&\,\Big(\frac{6}{11}\Big)^2\\[2mm] =&\,\frac{36}{121}, \end{align*}

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

\begin{align*} &\,P(T=14)\\[2mm] =&\,2\Big(\frac{6}{11}\Big)\\[2mm] &\,\hspace{2pt}\times\Big(\frac{2}{11}\Big)\\[2mm] &\,\hspace{4pt}+\Big(\frac3{11}\Big)^2\\[2mm] =&\,\frac{33}{121}, \end{align*}

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

and

\begin{align*} P(T=16)=&\,\Big(\frac{2}{11}\Big)^2\\[2mm] =&\,\frac{4}{121}. \end{align*}

Hence the sampling distribution is

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

The probabilities sum to

\begin{align*} &\,\frac{36+36+33+12+4}{121}\\[2mm] =&\,\frac{121}{121}=1, \end{align*}

as required.

(c)

解法一

思路

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按照 (b) 得到的抽样分布，将每个总数乘以它对应的概率，再把各项相加，得到期望。

答题过程

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Using the sampling distribution from part (b),

\begin{align*} &\,E(T)\\[2mm] =&\,12\Big(\frac{36}{121}\Big)\\[2mm] &\,\hspace{2pt}+13\Big(\frac{36}{121}\Big)\\[2mm] &\,\hspace{4pt}+14\Big(\frac{33}{121}\Big)\\[2mm] &\,\hspace{6pt}+15\Big(\frac{12}{121}\Big)\\[2mm] &\,\hspace{8pt}+16\Big(\frac4{121}\Big)\\[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\Big(\frac6{11}\Big)\\[2mm] &\,\hspace{2pt}+7\Big(\frac3{11}\Big)\\[2mm] &\,\hspace{4pt}+8\Big(\frac2{11}\Big)\\[2mm] =&\,\frac{73}{11}. \end{align*}

For two customers, linearity of expectation gives

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

S2 2022 Jan Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

解法二

思路

答题过程