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IAL 2020 Oct S2 Q3

A Level / Edexcel / S2

IAL 2020 Oct Paper · Question 3

题目

Problem

A manufacturer produces plates. The proportion of plates that are flawed is 45%, with flawed plates occurring independently.

A random sample of 10 of these plates is selected.

(a) Find the probability that the sample contains

(i) fewer than 2 flawed plates,

(ii) at least 6 flawed plates.

(4)

George believes that the proportion of flawed plates is not 45%. To assess his belief George takes a random sample of 120 plates. The random variable F represents the number of flawed plates found in the sample.

(b) Using a normal approximation, find the maximum number of plates, c, and the minimum number of plates, d, such that P(F ≤ c) ≤ 0.05 and P(F ≥ d) ≤ 0.05 where F ~ B(120, 0.45)

(7)

The manufacturer claims that, after a change to the production process, the proportion of flawed plates has decreased. A random sample of 30 plates, taken after the change to the production process, contains 8 flawed plates.

(c) Use a suitable hypothesis test, at the 5% level of significance, to assess the manufacturer’s claim. State your hypotheses clearly.

(4)

(Total for Question 3 is 15 marks)

题目中文翻译

一家制造商生产盘子。盘子有缺陷的比例为 45%,且缺陷盘子彼此独立出现。

随机抽取 10 个盘子。

(a) 求样本中:

(i) 少于 2 个有缺陷盘子的概率;

(ii) 至少 6 个有缺陷盘子的概率。

George 认为有缺陷盘子的比例不是 45%。为了检验他的看法,他抽取了 120 个盘子。随机变量 FF 表示样本中有缺陷盘子的数量。

(b) 使用正态近似,求最大的盘子数 cc 和最小的盘子数 dd,使得

P(Fc)0.05P(F \leqslant c) \leqslant 0.05P(Fd)0.05P(F \geqslant d) \leqslant 0.05

其中 FB(120,0.45)F \sim B(120,0.45)

制造商声称,在生产工艺改变后,有缺陷盘子的比例下降了。改变工艺后随机抽取 30 个盘子,其中 8 个有缺陷。

(c) 使用合适的假设检验,在 5% 显著性水平下检验制造商的说法,并清楚写出假设。

(第 3 题共 15 分)

解答

(a)(i)

解法一

思路

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每个盘子有缺陷的概率为 0.45,且各盘子彼此独立,所以缺陷盘子数服从二项分布。“少于 2 个”包括 0 个和 1 个,可直接计算累积概率。

答题过程

展开

Let XX be the number of flawed plates in the sample. Then

XB(10,0.45).X\sim B(10,0.45).

Therefore,

P(X<2)=P(X1)=(0.55)10+10(0.45)(0.55)9=0.0232571=0.0233.\begin{align*} P(X<2) =&\,P(X\leqslant1) \\[4mm] =&\,(0.55)^{10} +10(0.45)(0.55)^9 \\[4mm] =&\,0.0232571\ldots \\[4mm] =&\,\boxed{0.0233}. \end{align*}

(a)(ii)

解法一

思路

展开

“至少 6 个”是“不超过 5 个”的补集。沿用 (i) 的二项分布,计算 1P(X5)1-P(X\leqslant5)

答题过程

展开 P(X6)=1P(X5)=10.738437=0.261562=0.262.\begin{align*} P(X\geqslant6) =&\,1-P(X\leqslant5) \\[4mm] =&\,1-0.738437\ldots \\[4mm] =&\,0.261562\ldots \\[4mm] =&\,\boxed{0.262}. \end{align*}

(b)

解法一

思路

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用正态分布近似 B(120,0.45)B(120,0.45),其均值为 54、方差为 29.7。左右两尾各对应标准正态的 5% 分位点 ±1.64485\pm1.64485。计算 P(Fc)P(F\leqslant c) 时用边界 c+0.5c+0.5,计算 P(Fd)P(F\geqslant d) 时用边界 d0.5d-0.5,最后根据最大或最小整数要求取值。

答题过程

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The mean and variance are

μ=120(0.45)=54\mu=120(0.45)=54

and

σ2=120(0.45)(0.55)=29.7.\sigma^2=120(0.45)(0.55)=29.7.

Hence

F˙N(54,29.7).F\mathrel{\dot\sim}N(54,29.7).

For the lower tail, using a continuity correction,

c+0.55429.71.64485.\frac{c+0.5-54}{\sqrt{29.7}} \leqslant-1.64485.

Therefore,

c+0.5541.6448529.7=45.0359,\begin{align*} c+0.5 \leqslant&\,54-1.64485\sqrt{29.7} \\[4mm] =&\,45.0359\ldots, \end{align*}

so c44.5359c\leqslant44.5359\ldots. The maximum integer value is

c=44.\boxed{c=44}.

For the upper tail,

d0.55429.71.64485.\frac{d-0.5-54}{\sqrt{29.7}} \geqslant1.64485.

Therefore,

d0.554+1.6448529.7=62.9641,\begin{align*} d-0.5 \geqslant&\,54+1.64485\sqrt{29.7} \\[4mm] =&\,62.9641\ldots, \end{align*}

so d63.4641d\geqslant63.4641\ldots. The minimum integer value is

d=64.\boxed{d=64}.

(c)

解法一

思路

展开

制造商声称缺陷率下降,因此进行下尾单侧检验。以原比例 0.45 为原假设,计算在 B(30,0.45)B(30,0.45) 下出现 8 个或更少缺陷盘子的概率,并与 5% 比较。

答题过程

展开

Let pp be the proportion of flawed plates. The hypotheses are

H0:p=0.45,H1:p<0.45.H_0:p=0.45, \qquad H_1:p<0.45.

Under H0H_0, let

YB(30,0.45).Y\sim B(30,0.45).

The p-value is

P(Y8)=0.0312056P(Y\leqslant8)=0.0312056\ldots

Since 0.0312<0.050.0312<0.05, H0H_0 is rejected. There is sufficient evidence at the 5% significance level to support the manufacturer’s claim that the proportion of flawed plates has decreased.

解法二

思路

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也可以按官方评分资料接受的临界域路线作答。检查相邻累积概率可知,P(Y8)<0.05P(Y\leqslant8)<0.05,但 P(Y9)>0.05P(Y\leqslant9)>0.05,所以 5% 下尾临界域是 Y8Y\leqslant8

答题过程

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Use the hypotheses

H0:p=0.45,H1:p<0.45,H_0:p=0.45, \qquad H_1:p<0.45,

with YB(30,0.45)Y\sim B(30,0.45) under H0H_0. Now,

P(Y8)=0.03121<0.05,P(Y\leqslant8)=0.03121\ldots<0.05,

whereas

P(Y9)=0.06941>0.05.P(Y\leqslant9)=0.06941\ldots>0.05.

Hence the critical region is Y8Y\leqslant8. The observed value 88 lies in the critical region, so H0H_0 is rejected. The data support the manufacturer’s claim that the proportion of flawed plates has decreased.