在一个拥有 1000 名会员的社交俱乐部中，执行委员会希望调查会员对俱乐部所组织活动的意见。给出委员会宁愿获取意见样本而不询问所有会员的一个可能原因。解释如何获取一个容量为 100 的系统样本。给出在此背景下使用系统抽样的一个缺点。在社交俱乐部举办的一场活动中，十位朋友坐在一组特定的十个相邻座位上。确定其中两人（Adam 和 Bernice）之间至少隔着两人的可能座位排列数。

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完整解法

6((a))((i))

1. State a valid practical reason for choosing a sample over a census.

A sample is preferred because surveying the entire population of $1000$ members (a census) requires significantly more time, money, and administrative effort.

答案

It would be too time-consuming and costly to survey all $1000$ members.

6((a))((ii))

1. Determine the sampling interval $k$.

Population size, $N = 1000$
Sample size, $n = 100$
Sampling interval, $k = \frac{1000}{100} = 10$

2. Describe the full systematic sampling procedure.

Compile a numbered list of all $1000$ members.
Randomly select a starting number from $1$ to $10$ (for example, using a random number generator).
Select every $10$ th member on the list thereafter until exactly $100$ members are chosen.

答案

Number the members from $1$ to $1000$ in a list. Randomly select a starting number from $1$ to $10$ . Thereafter, select every $10$ th member on the list until a sample of $100$ is obtained.

6((a))((iii))

1. State a valid disadvantage of systematic sampling in the context of a membership list.

A key disadvantage of systematic sampling is its vulnerability to periodic or cyclical patterns in the sampling frame. If the club's membership list is structured such that similar types of members appear at regular intervals (for example, ordered by family units where every 10th person happens to be the head of a household), the resulting sample will not be representative of the entire club. Alternatively, one could note that unlike a simple random sample, not every possible group of 100 members has an equal chance of being selected (e.g., two members adjacent on the list can never be chosen together).

答案

If the membership list has a hidden periodic pattern (e.g. members are listed by couples or family units), picking every 10th person might consistently select people with similar demographics (e.g. only husbands), leading to a biased sample.

6((b))

1. Calculate the total number of unrestricted seating arrangements.

The total number of ways the 10 friends can sit in 10 seats without any restrictions is:
$10! = 3,628,800$

2. Calculate the number of arrangements where Adam and Bernice sit together (separated by 0 people).

Treat Adam and Bernice as a single, inseparable unit. There are now $9$ units to arrange (this 1 unit + the $8$ other friends).
The 9 units can be arranged in $9!$ ways, and Adam and Bernice can swap positions within their unit in $2!$ ways.
$\text{Number of arrangements} = 9! \times 2! = 725,760$

3. Calculate the number of arrangements where Adam and Bernice are separated by exactly 1 person.

Choose $1$ out of the $8$ other friends to sit exactly between Adam and Bernice ( $8$ possible choices).
Treat this block of 3 people as a single unit. Within this block, Adam and Bernice can swap places in $2!$ ways (the chosen friend remains in the middle).
There are now $8$ units to arrange in total (this new 3-person unit + the remaining $7$ friends).
$\text{Number of arrangements} = 8 \times 2! \times 8! = 645,120$

4. Subtract the complementary scenarios from the total to find the valid arrangements.

To find the number of ways they are separated by at least 2 people, subtract the number of ways they are separated by $0$ or $1$ person from the total unrestricted ways:
$\text{Valid arrangements} = 3,628,800 - 725,760 - 645,120$
$= 2,257,920$

答案

$2,257,920$

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完整解法

6((a))((i))

6((a))((ii))

6((a))((iii))

6((b))

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