题目

Problem

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

(1)

A factory produces beads in bags for craft shops. A small bag contains 40 beads, a medium bag contains 80 beads and a large bag contains 150 beads. The factory produces small, medium and large bags in the ratio 5:3:2 respectively.

A random sample of 3 bags is taken from the factory.

(b) Find the sampling distribution for the range of the number of beads in the 3 bags in the sample.

(7)

A random sample of n sets of 3 bags is taken. The random variable Y represents the number of these n sets of 3 bags that have a range of 70

(3)

(Total for Question 6 is 11 marks)

题目中文翻译

(a) 解释你对统计量的抽样分布的理解。

一家工厂为手工店生产袋装珠子。小袋含 40 颗珠子，中袋含 80 颗，大袋含 150 颗。工厂生产小、中、大袋的比例分别为 5:3:2。

随机抽取 3 袋。

(b) 求这 3 袋样本中珠子数量极差的抽样分布。

随机抽取 n 组，每组 3 袋。随机变量 $Y$ 表示这 n 组中极差为 70 的组数。

（第 6 题共 11 分）

解答

(a)

解法一

思路

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抽样分布描述的是某个统计量所有可能的取值，以及每个取值所对应的概率。

答题过程

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A sampling distribution is the probability distribution of a statistic: it gives all possible values of the statistic and their associated probabilities.

(b)

解法一

思路

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先由生产比例得到三种袋子的抽取概率。三个袋子的珠子数只能是 40、80、150，因此极差只可能是 0、40、70、110。极差为 0 时三个袋子同型；其余情形则列出能产生相应最小值与最大值的组合，并乘上排列数计算概率。

答题过程

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Let $R$ be the range. The probabilities of selecting the three bag sizes are

P(40)=0.5, \qquad P(80)=0.3, \qquad P(150)=0.2.

The possible values of $R$ are $0$ , $40$ , $70$ and $110$ . First,

\begin{align*} P(R=0) =&\,(0.5)^3+(0.3)^3+(0.2)^3 \\[4mm] =&\,0.16. \end{align*}

For a range of 40, the sample contains only 40s and 80s, with both sizes present. Thus,

\begin{align*} P(R=40) =&\,3(0.5)^2(0.3)+3(0.5)(0.3)^2 \\[4mm] =&\,0.36. \end{align*}

Similarly,

\begin{align*} P(R=70) =&\,3(0.3)^2(0.2)+3(0.3)(0.2)^2 \\[4mm] =&\,0.09, \end{align*}

and

\begin{align*} P(R=110) =&\,3(0.5)^2(0.2)+3(0.5)(0.2)^2 \\[2mm] &\,+6(0.5)(0.3)(0.2) \\[4mm] =&\,0.39. \end{align*}

Therefore, the sampling distribution of $R$ is

$r$	$0$	$40$	$70$	$110$
$P(R=r)$	$0.16$	$0.36$	$0.09$	$0.39$

(c)

解法一

思路

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由 (b) 可知，每组极差为 70 的概率是 0.09，所以 $Y$ 服从参数为 $n$ 和 0.09 的二项分布。事件 $Y=0$ 表示所有 $n$ 组的极差都不是 70，其概率为 $0.91^n$ ；解不等式后取满足条件的最小整数。

答题过程

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From part (b),

Y\sim\operatorname{B}(n,0.09).

Therefore,

P(Y=0)=(1-0.09)^n=(0.91)^n.

We require

(0.91)^n<0.2.

Taking logarithms gives

n>\frac{\log(0.2)}{\log(0.91)}=17.065\ldots

Hence the minimum value is

\boxed{n=18}.

IAL 2020 Oct S2 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