CalcGospel 國際數學圖譜
返回

S2 2020 Oct Q6

A Level / Edexcel / S2

IAL 2020 Oct Paper · Question 6

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 nn sets of 3 bags is taken. The random variable YY represents the number of these nn sets of 3 bags that have a range of 70

(c) Calculate the minimum value of nn such that P(Y=0)<0.2P(Y=0)<0.2

(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。

(1)

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。

(7)

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

(c) 求使 P(Y=0)<0.2P(Y=0)<0.2 的最小 nn。

(3)

（第 6 题共 11 分）

解答

(a)

解法一

思路

展开

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

展开

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

展开

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

展开

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

P(40)= 0.5,P(80)= 0.3,P(150)= 0.2.\begin{align*} P(40)=&\,0.5,\\[4mm] P(80)=&\,0.3,\\[4mm] P(150)=&\,0.2. \end{align*}

The possible values of RR are 00, 4040, 7070 and 110110. First,

P(R=0)P(R=0) = (0.5)3+(0.3)3 +(0.2)3= 0.16.\begin{align*} =&\,(0.5)^3+(0.3)^3\\ &\,\hspace{2pt}+(0.2)^3 \\[4mm] =&\,0.16. \end{align*}

For a range of 40, the possible combinations are (40,40,80)(40,40,80) and (40,80,80)(40,80,80). Each has 3!/2!=33!/2!=3 arrangements. Thus,

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

For a range of 70, the combinations are (80,80,150)(80,80,150) and (80,150,150)(80,150,150), each with 3 arrangements. Thus,

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

For a range of 110, the combinations are (40,40,150)(40,40,150), (40,150,150)(40,150,150) and (40,80,150)(40,80,150). The first two each have 3 arrangements, while the last has 3!=63!=6 arrangements. Hence,

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

Therefore, the sampling distribution of RR is

rrP(R=r)P(R=r)
000.160.16
40400.360.36
70700.090.09
1101100.390.39

(c)

解法一

思路

展开

由 (b) 可知，每组极差为 70 的概率是 0.09，所以 YY 服从参数为 nn 和 0.09 的二项分布。事件 Y=0Y=0 表示所有 nn 组的极差都不是 70，其概率为 0.91n0.91^n；取对数后，因 log⁡(0.91)<0\log(0.91)<0，除以它时不等号必须反向，再取满足条件的最小整数。

答题过程

展开

From part (b),

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

Therefore,

P(Y=0)= (1−0.09)n= (0.91)n.\begin{align*} P(Y=0)=&\,(1-0.09)^n\\[4mm] =&\,(0.91)^n. \end{align*}

We require

(0.91)n<0.2.(0.91)^n<0.2.

Taking logarithms gives

nlog⁡(0.91)<log⁡(0.2).n\log(0.91)<\log(0.2).

Since log⁡(0.91)<0\log(0.91)<0, dividing reverses the inequality:

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

Hence the minimum value is

n=18.\boxed{n=18}.