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CIE 9709 2024 March Paper 62 Q5

A Level / CIE / S2

CIE 9709 2024 March Paper 62 Paper · Question 5

题目

Problem

A teacher models the numbers of girls and boys who arrive late for her class on any day by the independent random variables G∼Po(0.10)G\sim\mathrm{Po}(0.10) and B∼Po(0.15)B\sim\mathrm{Po}(0.15) respectively.

(a) Find the probability that during a randomly chosen 22-day period no girls arrive late.

(1)

(b) Find the probability that during a randomly chosen 55-day period the total number of students who arrive late is less than 33.

(3)

(c) It is given that the values of P(G=r)\mathrm{P}(G=r) and P(B=r)\mathrm{P}(B=r) for r≥3r\geq 3 are very small and can be ignored.

Find the probability that on a randomly chosen day more girls arrive late than boys.

(3)

Following a timetable change the teacher claims that on average more students arrive late than before the change. During a randomly chosen 55-day period a total of 44 students are late.

(d) Test the teacher’s claim at the 5%5\% significance level.

(5)
题目中文翻译

一名老师分别用相互独立的随机变量 G∼Po(0.10)G\sim\mathrm{Po}(0.10) 和 B∼Po(0.15)B\sim\mathrm{Po}(0.15) 描述任意一天在她的课上迟到的女生人数和男生人数。

(a) 随机选取一段为期 22 天的时间，求没有女生迟到的概率。

(1)

(b) 随机选取一段为期 55 天的时间，求迟到的学生总人数少于 33 人的概率。

(3)

(c) 已知当 r≥3r\geq 3 时，P(G=r)\mathrm{P}(G=r) 和 P(B=r)\mathrm{P}(B=r) 的值很小，可以忽略。

随机选取一天，求迟到的女生人数多于迟到的男生人数的概率。

(3)

课程表调整后，这名老师声称迟到的学生人数平均比调整前多。在随机选取的一段为期 55 天的时间内，共有 44 名学生迟到。

(d) 在 5%5\% 的显著性水平下检验该老师的说法。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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两天女生迟到总数的泊松参数为0.2，取零次概率。

答题过程

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Over two days the number of girls arriving late has distribution Po(0.2)\mathrm{Po}(0.2). Hence

P(no girls late)=e−0.2=0.819(3 s.f.).\mathrm{P}(\text{no girls late})=e^{-0.2} =\boxed{0.819}\quad\text{(3 s.f.).}

(b)

解法一

思路

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五天两类人数相加的参数为1.25，少于3人即0、1、2人。

答题过程

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Let TT be the total number arriving late in five days. Then

T∼Po(5(0.10+0.15))=Po(1.25).T\sim\mathrm{Po}\bigl(5(0.10+0.15)\bigr)=\mathrm{Po}(1.25).

Thus

P(T<3)= e−1.25(1+1.25+1.2522!)≈ 0.868(3 s.f.).\begin{align*} \mathrm{P}(T<3)=&\,e^{-1.25}\bigg(1+1.25+\frac{1.25^2}{2!}\bigg)\\[4mm] \approx&\,\boxed{0.868}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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按要求忽略至少3人的事件后，女生多于男生只有(1,0)、(2,0)、(2,1)三种，分别用独立性求概率并相加。

答题过程

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Ignoring cases with G≥3G\geq3 or B≥3B\geq3, as instructed, the possible pairs (G,B)(G,B) with G>BG>B are

(1,0),(2,0),(2,1).(1,0),\qquad(2,0),\qquad(2,1).

By independence,

P(G>B)≈ P(G=1)P(B=0) +P(G=2)P(B=0) +P(G=2)P(B=1)= e−0.25(0.10+0.1022!+0.1022!(0.15))= 0.0823582…= 0.0824(3 s.f.).\begin{align*} \mathrm{P}(G>B)\approx&\,\mathrm{P}(G=1)\mathrm{P}(B=0)\\[4mm] &\,+\mathrm{P}(G=2)\mathrm{P}(B=0)\\[4mm] &\,+\mathrm{P}(G=2)\mathrm{P}(B=1)\\[4mm] =&\,e^{-0.25}\bigg(0.10+\frac{0.10^2}{2!} +\frac{0.10^2}{2!}(0.15)\bigg)\\[4mm] =&\,0.0823582\ldots\\[4mm] =&\,\boxed{0.0824}\quad\text{(3 s.f.).} \end{align*}

(d)

解法一

思路

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明确参数为每天平均迟到人数。原假设下五天总数服从Po1.25，观测4人采用至少4人的右尾检验。

答题过程

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Let λ\lambda be the daily mean number of students arriving late after the timetable change.

H0:λ=0.25,H1:λ>0.25.H_0:\lambda=0.25,\qquad H_1:\lambda>0.25.

Under H0H_0, the five-day total T∼Po(1.25)T\sim\mathrm{Po}(1.25). The upper-tail probability for the observed total of 4 is

 P(T≥4∣H0)= 1−e−1.25(1+1.25) −e−1.25(1.2522!+1.2533!)= 0.0382691….\begin{align*} &\,\mathrm{P}(T\geq4\mid H_0)\\[4mm] =&\,1-e^{-1.25}(1+1.25)\\[4mm] &\,-e^{-1.25}\bigg(\frac{1.25^2}{2!}+\frac{1.25^3}{3!}\bigg)\\[4mm] =&\,0.0382691\ldots. \end{align*}

Since 0.0382691…<0.050.0382691\ldots<0.05, we reject H0H_0. There is sufficient evidence at the 5% significance level to support the teacher’s claim that the mean number arriving late has increased.