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IAL 2021 June S2 Q2

A Level / Edexcel / S2

IAL 2021 June Paper · Question 2

题目

Problem

Luis makes and sells rugs. He knows that faults occur randomly in his rugs at a rate of 3 every 4 m^2

(a) Find the probability of there being exactly 5 faults in one of his rugs that is 4 m^2 in size.

(2)

(b) Find the probability that there are more than 5 faults in one of his rugs that is 6 m^2 in size.

(2)

Luis makes a rug that is 4 m^2 in size and finds it has exactly 5 faults in it.

(c) Write down the probability that the next rug that Luis makes, which is 4 m^2 in size, will have exactly 5 faults. Give a reason for your answer.

(2)

A small rug has dimensions 80 cm by 150 cm. Faults still occur randomly at a rate of 3 every 4 m^2

Luis makes a profit of £80 on each small rug he sells that contains no faults but a profit of £60 on any small rug he sells that contains faults.

Luis sells n small rugs and expects to make a profit of at least £4000

(d) Calculate the minimum value of n

(4)

Luis wishes to increase the productivity of his business and employs Rhiannon. Faults also occur randomly in Rhiannon’s rugs and independently to faults made by Luis.

Luis randomly selects 10 small rugs made by Rhiannon and finds 13 faults.

(e) Test, at the 5% level of significance, whether or not there is evidence to support the suggestion that the rate at which faults occur is higher for Rhiannon than for Luis. State your hypotheses clearly.

(5)

(Total for Question 2 is 15 marks)

题目中文翻译

Luis 制作并销售地毯。他知道自己地毯上的瑕疵是随机出现的，速率为每 4 m^2 有 3 个瑕疵。

(a) 求一块 4 m^2 地毯上恰好有 5 个瑕疵的概率。

(b) 求一块 6 m^2 地毯上瑕疵多于 5 个的概率。

Luis 做了一块 4 m^2 的地毯，发现其中恰好有 5 个瑕疵。

(c) 写下 Luis 接下来做的一块 4 m^2 地毯恰好有 5 个瑕疵的概率，并给出理由。

一块小地毯的尺寸为 80 cm × 150 cm。瑕疵仍以每 4 m^2 有 3 个的速率随机出现。

Luis 每卖出一块没有瑕疵的小地毯可赚 £80；每卖出一块有瑕疵的小地毯可赚 £60。

Luis 卖出 n 块小地毯，并期望总利润至少为 £4000。

(d) 求 n 的最小值。

Luis 想提高业务效率，于是雇佣了 Rhiannon。Rhiannon 的地毯也随机出现瑕疵，而且与 Luis 的瑕疵相互独立。

Luis 随机抽查了 Rhiannon 制作的 10 块小地毯，发现共有 13 个瑕疵。

(e) 在 5% 显著性水平下检验是否有证据支持“Rhiannon 的瑕疵发生率高于 Luis”的说法，并清楚写出假设。

（第 2 题共 15 分）

解答

(a)

解法一

思路

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瑕疵随机且以固定面积速率出现，因此使用泊松分布。4 平方米地毯的平均瑕疵数为 3，直接计算恰好 5 个瑕疵的概率。

答题过程

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Let XX be the number of faults in a 4 m24\text{ m}^2 rug. Then

X∼Po⁡(3).X\sim\operatorname{Po}(3).

Therefore,

P(X=5)= e−3355!= 0.100818…= 0.101(3 s.f.).\begin{align*} P(X=5) =&\,\frac{e^{-3}3^5}{5!} \\[4mm] =&\,0.100818\ldots \\[4mm] =&\,\boxed{0.101}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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面积从 4 平方米增至 6 平方米，泊松均值按面积同比例变为 3×6/4=4.53\times6/4=4.5。“多于 5 个”使用补事件“不超过 5 个”计算。

答题过程

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Let YY be the number of faults in a 6 m26\text{ m}^2 rug. Then

Y∼Po⁡(4.5).Y\sim\operatorname{Po}(4.5).

Hence,

P(Y>5)= 1−P(Y⩽5)= 1−0.702930…= 0.297069…= 0.297(3 s.f.).\begin{align*} P(Y>5) =&\,1-P(Y\leqslant5) \\[4mm] =&\,1-0.702930\ldots \\[4mm] =&\,0.297069\ldots \\[4mm] =&\,\boxed{0.297}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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泊松过程具有独立增量，前一块地毯出现 5 个瑕疵不会改变下一块同面积地毯的分布。因此所求概率与 (a) 完全相同。

答题过程

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The probability is

0.101.\boxed{0.101}.

Faults occur randomly and independently, so the number of faults in the previous rug does not affect the number in the next rug.

(d)

解法一

思路

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先把小地毯面积换算为 1.2 m21.2\text{ m}^2，所以每块的泊松均值为 0.90.9。无瑕疵概率为 e−0.9e^{-0.9}；据此计算每块地毯的期望利润，再令 nn 块的期望总利润至少为 £4000，最后向上取整。

答题过程

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The area of one small rug is

0.8×1.5=1.2 m2.0.8\times1.5=1.2\text{ m}^2.

If FF is the number of faults in one small rug, then

F∼Po⁡(3×1.24)=Po⁡(0.9).F\sim\operatorname{Po}\left(3\times\frac{1.2}{4}\right) =\operatorname{Po}(0.9).

Thus

P(F=0)=e−0.9.P(F=0)=e^{-0.9}.

The expected profit from one small rug is

E(profit)= 80e−0.9+60(1−e−0.9)= 60+20e−0.9.\begin{align*} E(\text{profit}) =&\,80e^{-0.9}+60(1-e^{-0.9}) \\[4mm] =&\,60+20e^{-0.9}. \end{align*}

Therefore,

n(60+20e−0.9)⩾ 4000n⩾ 58.71….\begin{align*} n(60+20e^{-0.9})\geqslant&\,4000 \\[4mm] n\geqslant&\,58.71\ldots. \end{align*}

Since nn is a whole number, its minimum value is

n=59.\boxed{n=59}.

(e)

解法一

思路

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10 块小地毯总面积为 12 平方米；若 Rhiannon 与 Luis 的瑕疵率相同，总瑕疵数的泊松均值应为 9。“更高”对应右尾检验，计算至少观察到 13 个瑕疵的概率并与 0.050.05 比较。

答题过程

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Let λ\lambda be the mean number of faults in 10 small rugs made by Rhiannon. Test

H0:λ=9againstH1:λ>9.H_0:\lambda=9 \qquad\text{against}\qquad H_1:\lambda>9.

Under H0H_0, let RR be the total number of faults. Then

R∼Po⁡(9).R\sim\operatorname{Po}(9).

The pp-value is

P(R⩾13)= 1−P(R⩽12)= 0.1242….\begin{align*} P(R\geqslant13) =&\,1-P(R\leqslant12) \\[4mm] =&\,0.1242\ldots. \end{align*}

Since 0.1242>0.050.1242>0.05, do not reject H0H_0. There is insufficient evidence at the 5% level of significance that the rate at which faults occur is higher for Rhiannon than for Luis.

解法二

思路

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也可以用临界域完成右尾检验。比较相邻的右尾概率，找到概率首次不超过 0.050.05 的最小瑕疵数，再判断观测值 13 是否进入临界域。

答题过程

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Use the same hypotheses and R∼Po⁡(9)R\sim\operatorname{Po}(9) under H0H_0. Now,

P(R⩾14)=0.0739P(R\geqslant14)=0.0739

and

P(R⩾15)=0.0415.P(R\geqslant15)=0.0415.

Therefore the 5% upper-tail critical region is

R⩾15.R\geqslant15.

The observed value, 1313, is not in the critical region, so do not reject H0H_0. There is insufficient evidence at the 5% level of significance that the rate at which faults occur is higher for Rhiannon than for Luis.