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S2 2020 Oct Q4

A Level / Edexcel / S2

IAL 2020 Oct Paper · Question 4

题目

Problem

In a peat bog, Common Spotted-orchids occur at a mean rate of 4.5 per m2\text{m}^2

(a) Give an assumption, not already stated, that is required for the number of Common Spotted-orchids per m2\text{m}^2 of the peat bog to follow a Poisson distribution.

(1)

Given that the number of Common Spotted-orchids in 1 m2\text{m}^2 of the peat bog can be modelled by a Poisson distribution,

(b) find the probability that in a randomly selected 1 m2\text{m}^2 of the peat bog

(i) there are exactly 6 Common Spotted-orchids,

(ii) there are fewer than 10 but more than 4 Common Spotted-orchids.

(4)

Juan believes that by introducing a new management scheme the number of Common Spotted-orchids in the peat bog will increase. After three years under the new management scheme, a randomly selected 2 m2\text{m}^2 of the peat bog contains 11 Common Spotted-orchids.

(c) Using a 5% significance level assess Juan’s belief. State your hypotheses clearly.

(5)

Assuming that in the peat bog, Common Spotted-orchids still occur at a mean rate of 4.5 per m2\text{m}^2

(d) use a normal approximation to find the probability that in a randomly selected 20 m2\text{m}^2 of the peat bog there are fewer than 70 Common Spotted-orchids.

(3)

Following a period of dry weather, the probability that there are fewer than 70 Common Spotted-orchids in a randomly selected 20 m2\text{m}^2 of the peat bog is 0.012

A random sample of 200 non-overlapping 20 m2\text{m}^2 areas of the peat bog is taken.

(e) Using a suitable approximation, calculate the probability that at most 1 of these areas contains fewer than 70 Common Spotted-orchids.

(3)

(Total for Question 4 is 16 marks)

题目中文翻译

在一个泥炭沼泽中,斑点兰的平均出现率为每平方米 4.5 株。

(a) 写出一条未在题干中说明、但使每平方米斑点兰数量服从泊松分布所需的假设。

(1)

已知该泥炭沼泽中 1 m2\text{m}^2 的斑点兰数量可用泊松分布建模。

(b) 求随机选取的 1 m2\text{m}^2 沼泽中:

(i) 恰好有 6 株斑点兰的概率;

(ii) 多于 4 株但少于 10 株斑点兰的概率。

(4)

Juan 认为通过引入新的管理方案,沼泽中的斑点兰数量会增加。经过三年新的管理方案后,随机选取的 2 m2\text{m}^2 沼泽中有 11 株斑点兰。

(c) 在 5% 显著性水平下检验 Juan 的看法,并清楚写出假设。

(5)

假设泥炭沼泽中的斑点兰仍以每平方米 4.5 株的平均速率出现。

(d) 使用正态近似,求随机选取的 20 m2\text{m}^2 沼泽中斑点兰少于 70 株的概率。

(3)

在一段干旱天气后,随机选取的 20 m2\text{m}^2 沼泽中斑点兰少于 70 株的概率为 0.012。

随机抽取 200 个互不重叠的 20 m2\text{m}^2 区域。

(e) 使用合适的近似,求其中至多 1 个区域含有少于 70 株斑点兰的概率。

(3)

(第 4 题共 16 分)

解答

(a)

解法一

思路

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泊松模型要求事件在区域内随机且彼此独立地发生。题目已经给出恒定的平均出现率,因此补充独立性这一假设即可。

答题过程

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Common Spotted-orchids occur independently of one another.

(b)(i)

解法一

思路

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在 1 m21\text{ m}^2 区域内,斑点兰数量服从均值为 4.5 的泊松分布。把 s=6s=6 代入泊松概率质量函数。

答题过程

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Let SS be the number of Common Spotted-orchids in 1 m21\text{ m}^2. Then

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

Therefore,

P(S=6)= e−4.5(4.5)66!= 0.128120…= 0.128.\begin{align*} P(S=6) =&\,\frac{e^{-4.5}(4.5)^6}{6!} \\[4mm] =&\,0.128120\ldots \\[4mm] =&\,\boxed{0.128}. \end{align*}

解法二

思路

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累积到 6 的概率减去累积到 5 的概率,剩下的就是恰好为 6 的概率。

答题过程

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For S∼Po⁡(4.5)S\sim\operatorname{Po}(4.5),

P(S=6)P(S=6) = P(S⩽6)−P(S⩽5)= 0.831050… −0.702930…= 0.128120…= 0.128.\begin{align*} =&\,P(S\leqslant6)-P(S\leqslant5) \\[4mm] =&\,0.831050\ldots \\ &\,{}-0.702930\ldots \\[4mm] =&\,0.128120\ldots \\[4mm] =&\,\boxed{0.128}. \end{align*}

(b)(ii)

解法一

思路

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“多于 4 但少于 10”对应 5⩽S⩽95\leqslant S\leqslant9。利用两个累积概率之差 P(S⩽9)−P(S⩽4)P(S\leqslant9)-P(S\leqslant4) 计算。

答题过程

展开 P(4<S<10)P(4<S<10) = P(S⩽9)−P(S⩽4)= 0.9829…−0.5321…= 0.450803…= 0.451.\begin{align*} =&\,P(S\leqslant9)-P(S\leqslant4) \\[4mm] =&\,0.9829\ldots-0.5321\ldots \\[4mm] =&\,0.450803\ldots \\[4mm] =&\,\boxed{0.451}. \end{align*}

(c)

解法一

思路

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Juan 认为平均数量增加,因此进行上尾单侧检验。原平均率为每平方米 4.5 株,所以在 2 m22\text{ m}^2 内原假设的均值是 9。计算观测到至少 11 株的上尾概率,并与 5% 比较。

答题过程

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Let λ\lambda be the mean number of Common Spotted-orchids in 2 m22\text{ m}^2. The hypotheses are

H0:λ=9,H1:λ>9.\begin{gathered} H_0:\lambda=9,\\[2mm] H_1:\lambda>9 \end{gathered}.

Under H0H_0, let

M∼Po⁡(9).M\sim\operatorname{Po}(9).

The p-value is

P(M⩾11)P(M\geqslant11) = 1−P(M⩽10)= 0.294011…\begin{align*} =&\,1-P(M\leqslant10) \\[4mm] =&\,0.294011\ldots \end{align*}

Since 0.294>0.050.294>0.05, H0H_0 is not rejected. There is insufficient evidence at the 5% significance level to support Juan’s belief that the number of Common Spotted-orchids has increased.

解法二

思路

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也可以使用临界域路线。由上尾概率找出不超过 5% 的最小临界值:P(M⩾15)<0.05P(M\geqslant15)<0.05,但把边界扩到 14 后概率会超过 5%,所以临界域是 M⩾15M\geqslant15。

答题过程

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Use the hypotheses

H0:λ=9,H1:λ>9,\begin{gathered} H_0:\lambda=9,\\[2mm] H_1:\lambda>9 \end{gathered},

with M∼Po⁡(9)M\sim\operatorname{Po}(9) under H0H_0. Now,

P(M⩾15)= 0.041466…< 0.05,\begin{align*} P(M\geqslant15) =&\,0.041466\ldots \\[4mm] <&\,0.05, \end{align*}

whereas

P(M⩾14)= 0.073850…> 0.05.\begin{align*} P(M\geqslant14) =&\,0.073850\ldots \\[4mm] >&\,0.05. \end{align*}

Hence the critical region is M⩾15M\geqslant15. Since the observed value 1111 is not in the critical region, H0H_0 is not rejected. There is insufficient evidence to support Juan’s belief.

(d)

解法一

思路

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在 20 m220\text{ m}^2 内,泊松均值为 20×4.5=9020\times4.5=90;用均值和方差同为 90 的正态分布近似。“少于 70”即不超过 69,连续性修正后的边界是 69.5。

答题过程

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Let TT be the number of Common Spotted-orchids in 20 m220\text{ m}^2. Using a normal approximation,

T∼˙N(90,90).T\mathrel{\dot\sim}N(90,90).

Applying a continuity correction,

P(T<70)P(T<70) ≈ P(Z<69.5−9090)= P(Z<−2.160…)= 0.015386…= 0.0154.\begin{align*} \approx&\,P\Big(Z<\frac{69.5-90}{\sqrt{90}}\Big) \\[4mm] =&\,P(Z<-2.160\ldots) \\[4mm] =&\,0.015386\ldots \\[4mm] =&\,\boxed{0.0154}. \end{align*}

(e)

解法一

思路

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设符合条件的区域数为 VV,则原本有 V∼B(200,0.012)V\sim B(200,0.012)。因为样本量大而成功概率小,用均值 200×0.012=2.4200\times0.012=2.4 的泊松分布近似,再计算 P(V⩽1)P(V\leqslant1)。

答题过程

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Let VV be the number of areas containing fewer than 70 Common Spotted-orchids. Then

V∼B(200,0.012).V\sim B(200,0.012).

Using a Poisson approximation,

λ=200×0.012=2.4.\lambda=200\times0.012=2.4. V∼˙Po⁡(2.4).V\mathrel{\dot\sim}\operatorname{Po}(2.4).

Therefore,

P(V⩽1)P(V\leqslant1) = P(V=0)+P(V=1)= e−2.4+2.4e−2.4= 0.308441…= 0.308.\begin{align*} =&\,P(V=0)+P(V=1) \\[4mm] =&\,e^{-2.4}+2.4e^{-2.4} \\[4mm] =&\,0.308441\ldots \\[4mm] =&\,\boxed{0.308}. \end{align*}