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IAL 2021 Jan S2 Q1

A Level / Edexcel / S2

IAL 2021 Jan Paper · Question 1

题目

Problem

Jim farms oysters in a particular lake. He knows from past experience that 5% of young oysters do not survive to be harvested.

In a random sample of 30 young oysters, the random variable X represents the number that do not survive to be harvested.

(a) Write down a suitable model for the distribution of X.

(1)

(b) State an assumption that has been made for the model in part (a).

(1)

(c) Find the probability that

(i) exactly 24 young oysters do survive to be harvested,

(ii) at least 3 young oysters do not survive to be harvested.

(4)

A second random sample, of 200 young oysters, is taken.

The probability that at least n of these young oysters do not survive to be harvested is more than 0.8

(d) Using a suitable approximation, find the maximum value of n.

(3)

Jim believes that the level of salt in the lake water has changed and it has altered the survival rate of his oysters. He takes a random sample of 25 young oysters and places them in the lake.

When Jim harvests the oysters, he finds that 21 do survive to be harvested.

(e) Use a suitable test, at the 5% level of significance, to assess whether or not there is evidence that the proportion of oysters not surviving to be harvested is more than 5%. State your hypotheses clearly.

(5)

(Total for Question 1 is 14 marks)

题目中文翻译

Jim 在某个湖里养殖牡蛎。他根据以往经验知道，年轻牡蛎中有 5% 无法存活到收获。

在一个随机抽取的 30 只年轻牡蛎样本中，随机变量 XX 表示无法存活到收获的数量。

(a) 写下 XX 分布的合适模型。

(b) 写出 (a) 中模型所作的一个假设。

(c) 求以下概率：

(i) 恰好 24 只年轻牡蛎能够存活到收获；

(ii) 至少 3 只年轻牡蛎无法存活到收获。

再随机抽取 200 只年轻牡蛎。

至少有 nn 只无法存活到收获的概率大于 0.8。

(d) 使用合适的近似，求 nn 的最大值。

Jim 认为湖水盐度发生了变化，从而改变了牡蛎的存活率。他随机取 25 只年轻牡蛎放入湖中。

当他收获这些牡蛎时，发现有 21 只能够存活。

(e) 使用合适的检验，在 5% 显著性水平下判断是否有证据表明无法存活到收获的牡蛎比例大于 5%，并清楚写出假设。

（第 1 题共 14 分）

解答

(a)

解法一

思路

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样本大小固定为 30，每只牡蛎只有“存活”或“未存活”两种结果，而 XX 统计未存活的数量；单只未存活的概率为 0.050.05，因此使用二项分布。

答题过程

展开 X∼B⁡(30,0.05).\boxed{X\sim\operatorname{B}(30,0.05)}.

(b)

解法一

思路

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二项模型要求每次试验的成功概率保持不变；结合本题语境，可假设每只牡蛎无法存活到收获的概率均为 0.050.05。

答题过程

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The probability that an oyster does not survive to be harvested is constant for every oyster.

解法二

思路

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官方评分资料也接受独立性假设：一只牡蛎是否存活，不会影响其他牡蛎是否存活。

答题过程

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Whether each oyster survives to be harvested is independent of whether the other oysters survive.

(c)(i)

解法一

思路

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恰好 24 只存活等价于 30 只中恰好 6 只未存活。直接使用 (a) 中未存活数量的二项模型计算。

答题过程

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Exactly 24 survive if exactly 6 do not survive. Therefore,

P(X=6)= (306)(0.05)6(0.95)24= 0.00270899…= 0.00271(3 s.f.).\begin{align*} P(X=6) =&\,{30\choose6}(0.05)^6(0.95)^{24} \\[4mm] =&\,0.00270899\ldots \\[4mm] =&\,\boxed{0.00271}\quad\text{(3 s.f.)}. \end{align*}

(c)(ii)

解法一

思路

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“至少 3 只未存活”包含许多取值，使用补事件“不超过 2 只未存活”可以减少计算量。

答题过程

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Using X∼B⁡(30,0.05)X\sim\operatorname{B}(30,0.05),

P(X⩾3)= 1−P(X⩽2)= 1−0.812178…= 0.187821…= 0.188(3 s.f.).\begin{align*} P(X\geqslant3) =&\,1-P(X\leqslant2) \\[4mm] =&\,1-0.812178\ldots \\[4mm] =&\,0.187821\ldots \\[4mm] =&\,\boxed{0.188}\quad\text{(3 s.f.)}. \end{align*}

解法二

思路

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也可以改为统计存活数量。至少 3 只未存活等价于至多 27 只存活，再用存活概率 0.950.95 的二项分布求下尾概率。

答题过程

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Let SS be the number that survive. Then

S∼B⁡(30,0.95).S\sim\operatorname{B}(30,0.95).

Hence,

P(X⩾3)=P(S⩽27)=0.188(3 s.f.).P(X\geqslant3)=P(S\leqslant27) =\boxed{0.188}\quad\text{(3 s.f.)}.

(d)

解法一

思路

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200 只牡蛎中未存活的数量服从 B⁡(200,0.05)\operatorname{B}(200,0.05)。因为样本量大而概率较小，用均值为 200×0.05=10200\times0.05=10 的泊松分布近似。比较 n=7n=7 与下一个整数 n=8n=8 的上尾概率，才能确认最大值。

答题过程

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Let AA be the number that do not survive in the second sample. Then

A∼B⁡(200,0.05)≈Po⁡(10).A\sim\operatorname{B}(200,0.05) \approx\operatorname{Po}(10).

For n=7n=7,

P(A⩾7)= 1−P(A⩽6)= 1−0.130141…= 0.869859…>0.8.\begin{align*} P(A\geqslant7) =&\,1-P(A\leqslant6) \\[4mm] =&\,1-0.130141\ldots \\[4mm] =&\,0.869859\ldots>0.8. \end{align*}

For the next integer, n=8n=8,

P(A⩾8)= 1−P(A⩽7)= 0.779779…<0.8.\begin{align*} P(A\geqslant8) =&\,1-P(A\leqslant7) \\[4mm] =&\,0.779779\ldots<0.8. \end{align*}

Therefore, the maximum value is

n=7.\boxed{n=7}.

解法二

思路

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官方评分资料也接受正态近似。二项分布的均值为 10、方差为 9.5；把右尾概率大于 0.80.8 转成左尾概率小于 0.20.2，使用标准正态分位数求出 nn 的范围。

答题过程

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Alternatively,

A≈N(10,9.5).A\approx N(10,9.5).

The condition P(A⩾n)>0.8P(A\geqslant n)>0.8 gives

P(A<n)<0.2.P(A<n)<0.2.

Using Φ(−0.8416)=0.2\Phi(-0.8416)=0.2,

n−109.5< −0.8416n< 7.406….\begin{align*} \frac{n-10}{\sqrt{9.5}}<&\,-0.8416 \\[4mm] n<&\,7.406\ldots. \end{align*}

Hence the maximum integer value is

n=7.\boxed{n=7}.

(e)

解法一

思路

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样本中有 25−21=425-21=4 只未存活。题目问未存活比例是否高于 5%5\%，所以进行右尾检验；在原假设下计算至少 4 只未存活的概率，并与 0.050.05 比较。

答题过程

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Let pp be the proportion of oysters that do not survive. Test

H0:p=0.05againstH1:p>0.05.H_0:p=0.05 \qquad\text{against}\qquad H_1:p>0.05.

Under H0H_0, let CC be the number that do not survive. Then

C∼B⁡(25,0.05).C\sim\operatorname{B}(25,0.05).

Since 4 oysters do not survive, the pp-value is

P(C⩾4)= 1−P(C⩽3)= 0.0340906….\begin{align*} P(C\geqslant4) =&\,1-P(C\leqslant3) \\[4mm] =&\,0.0340906\ldots. \end{align*}

Since 0.0341<0.050.0341<0.05, reject H0H_0. There is evidence at the 5% level of significance that the proportion of oysters not surviving to be harvested is greater than 5%.

解法二

思路

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也可以统计存活数量并使用临界域。此时“未存活比例增加”等价于“存活比例下降”，所以备择假设及临界域都位于左尾。

答题过程

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Let qq be the proportion of oysters that survive. Test

H0:q=0.95againstH1:q<0.95.H_0:q=0.95 \qquad\text{against}\qquad H_1:q<0.95.

Under H0H_0, let DD be the number that survive. Then

D∼B⁡(25,0.95).D\sim\operatorname{B}(25,0.95).

Now,

P(D⩽21)=0.0341P(D\leqslant21)=0.0341

whereas

P(D⩽22)=0.1271.P(D\leqslant22)=0.1271.

Therefore the 5% lower-tail critical region is D⩽21D\leqslant21. The observed value, 2121, lies in the critical region, so reject H0H_0. There is evidence at the 5% level of significance that the proportion of oysters not surviving to be harvested is greater than 5%.