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IAL 2021 June S2 Q1

A Level / Edexcel / S2

IAL 2021 June Paper · Question 1

题目

Problem

Spany sells seeds and claims that 5% of its pansy seeds do not germinate. A packet of pansy seeds contains 20 seeds. Each seed germinates independently of the other seeds.

(a) Find the probability that in a packet of Spany’s pansy seeds

(i) more than 2 but fewer than 5 seeds do not germinate,

(ii) more than 18 seeds germinate.

(5)

Jem buys 5 packets of Spany’s pansy seeds.

(b) Calculate the probability that all of these packets contain more than 18 seeds that germinate.

(2)

Jem believes that Spany’s claim is incorrect. She believes that the percentage of pansy seeds that do not germinate is greater than 5%

(c) Write down the hypotheses for a suitable test to examine Jem’s belief.

(1)

Jem planted all of the 100 seeds she bought from Spany and found that 8 did not germinate.

(d) Using a suitable approximation, carry out the test using a 5% level of significance.

(6)

(Total for Question 1 is 14 marks)

题目中文翻译

Spany 销售种子，并声称其三色堇种子中有 5% 不发芽。每包三色堇种子有 20 粒，并且每粒种子是否发芽相互独立。

(a) 求在一包 Spany 的三色堇种子中：

(i) 不发芽的粒数多于 2 但少于 5 的概率；

(ii) 发芽的粒数多于 18 的概率。

Jem 买了 5 包 Spany 的三色堇种子。

(b) 计算这 5 包种子都含有多于 18 粒发芽种子的概率。

Jem 认为 Spany 的说法不正确，她认为不发芽的三色堇种子百分比大于 5%。

(c) 写下一个合适检验所需的假设。

Jem 把她从 Spany 买来的 100 粒种子全部种下，发现有 8 粒没有发芽。

(d) 使用适当的近似，在 5% 显著性水平下进行检验。

（第 1 题共 14 分）

解答

(a)(i)

解法一

思路

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令 XX 表示一包中不发芽的种子数，则 XX 服从参数为 2020、0.050.05 的二项分布。“多于 2 但少于 5”只包括 X=3X=3 和 X=4X=4。

答题过程

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Let XX be the number of seeds in a packet that do not germinate. Then

X∼B⁡(20,0.05).X\sim\operatorname{B}(20,0.05).

Therefore,

P(2<X<5)= P(X=3)+P(X=4)= (203)(0.05)3(0.95)17 +(204)(0.05)4(0.95)16= 0.072909…= 0.0729(3 s.f.).\begin{align*} P(2<X<5) =&\,P(X=3)+P(X=4) \\[4mm] =&\,{20\choose3}(0.05)^3(0.95)^{17} \\[4mm] &\,\hspace{2pt}+{20\choose4}(0.05)^4(0.95)^{16} \\[4mm] =&\,0.072909\ldots \\[4mm] =&\,\boxed{0.0729}\quad\text{(3 s.f.)}. \end{align*}

(a)(ii)

解法一

思路

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多于 18 粒发芽表示有 19 或 20 粒发芽，等价于不发芽的种子不超过 1 粒。继续使用 (a)(i) 的 XX，直接计算二项分布的下尾概率。

答题过程

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More than 18 seeds germinate if at most 1 seed does not germinate. Hence,

P(X⩽1)= P(X=0)+P(X=1)= (0.95)20+(201)(0.05)(0.95)19= 0.735839…= 0.736(3 s.f.).\begin{align*} P(X\leqslant1) =&\,P(X=0)+P(X=1) \\[4mm] =&\,(0.95)^{20} +{20\choose1}(0.05)(0.95)^{19} \\[4mm] =&\,0.735839\ldots \\[4mm] =&\,\boxed{0.736}\quad\text{(3 s.f.)}. \end{align*}

解法二

思路

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官方评分资料也接受直接统计发芽种子数。此时成功概率为 0.950.95，所求为恰好 19 粒或 20 粒发芽。

答题过程

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Let GG be the number of seeds that germinate. Then

G∼B⁡(20,0.95).G\sim\operatorname{B}(20,0.95).

Thus,

P(G>18)= P(G=19)+P(G=20)= (2019)(0.95)19(0.05) +(0.95)20= 0.736(3 s.f.).\begin{align*} P(G>18) =&\,P(G=19)+P(G=20) \\[4mm] =&\,{20\choose19}(0.95)^{19}(0.05) \\[4mm] &\,\hspace{2pt}+(0.95)^{20} \\[4mm] =&\,\boxed{0.736}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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每包中多于 18 粒发芽的概率由 (a)(ii) 求得。5 包彼此独立，因此 5 包全部满足条件的概率是单包概率的五次方；计算时使用未舍入值。

答题过程

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Using the unrounded probability from part (a)(ii),

P(all 5 packets)= (0.735839…)5= 0.21573…= 0.216(3 s.f.).\begin{align*} P(\text{all 5 packets}) =&\,(0.735839\ldots)^5 \\[4mm] =&\,0.21573\ldots \\[4mm] =&\,\boxed{0.216}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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令 pp 表示种子不发芽的比例。Spany 的说法放在原假设中；Jem 认为比例更高，所以备择假设是右尾的 p>0.05p>0.05。

答题过程

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Let pp be the proportion of pansy seeds that do not germinate. Then

H0:p=0.05andH1:p>0.05.\boxed{H_0:p=0.05 \qquad\text{and}\qquad H_1:p>0.05}.

(d)

解法一

思路

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在原假设下，100 粒中不发芽的数量服从 B⁡(100,0.05)\operatorname{B}(100,0.05)。由于样本量大而不发芽概率小，使用均值为 5 的泊松近似；观测到 8 粒不发芽，计算右尾概率并与 0.050.05 比较。

答题过程

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Let VV be the number of seeds, out of 100, that do not germinate. Under H0H_0,

V∼B⁡(100,0.05)≈Po⁡(5).V\sim\operatorname{B}(100,0.05) \approx\operatorname{Po}(5).

The approximate pp-value is

P(V⩾8)= 1−P(V⩽7)= 1−0.8666…= 0.1334….\begin{align*} P(V\geqslant8) =&\,1-P(V\leqslant7) \\[4mm] =&\,1-0.8666\ldots \\[4mm] =&\,0.1334\ldots. \end{align*}

Since 0.1334>0.050.1334>0.05, do not reject H0H_0. There is insufficient evidence at the 5% level of significance that the percentage of Spany’s pansy seeds that do not germinate is greater than 5%.

解法二

思路

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仍用均值为 5 的泊松近似，但改用右尾临界域。比较相邻两个右尾概率，确定累计概率首次不超过 0.050.05 的最小不发芽粒数，再判断观测值 8 是否落入其中。

答题过程

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Under H0H_0, use V∼Po⁡(5)V\sim\operatorname{Po}(5). Now,

P(V⩾9)=0.0681P(V\geqslant9)=0.0681

and

P(V⩾10)=0.0318.P(V\geqslant10)=0.0318.

Therefore the 5% upper-tail critical region is

V⩾10.V\geqslant10.

The observed value, 88, is not in the critical region, so do not reject H0H_0. There is insufficient evidence at the 5% level of significance that the percentage of Spany’s pansy seeds that do not germinate is greater than 5%.

解法三

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 5、方差为 100(0.05)(0.95)=4.75100(0.05)(0.95)=4.75；对“至少 8 粒”使用 7.57.5 的连续性修正，再求右尾概率。

答题过程

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Alternatively, under H0H_0, let

Z∼N(5,4.75)Z\sim N(5,4.75)

approximate the distribution of VV.

Applying the continuity correction,

P(V⩾8)≈P(Z>7.5)=1−Φ(7.5−54.75)=0.126(3 s.f.).\begin{align*} P(V\geqslant8) &\approx P(Z>7.5) \\[4mm] &=1-\Phi\left(\frac{7.5-5}{\sqrt{4.75}}\right) \\[4mm] &=0.126\quad\text{(3 s.f.)}. \end{align*}

Since 0.126>0.050.126>0.05, do not reject H0H_0. There is insufficient evidence at the 5% level of significance that the percentage of Spany’s pansy seeds that do not germinate is greater than 5%.