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CIE 9709 2023 November Paper 63 Q3

A Level / CIE / S2

CIE 9709 2023 Nov Paper 63 Paper · Question 3

题目

Problem

A website owner finds that, on average, his website receives 0.30.3 hits per minute. He believes that the number of hits per minute follows a Poisson distribution.

(a) Assume that the owner is correct.

(i) Find the probability that there will be at least 44 hits during a 1010-minute period.

(3)

(ii) Use a suitable approximating distribution to find the probability that there will be fewer than 4040 hits during a 33-hour period.

(4)

A friend agrees that the website receives, on average, 0.30.3 hits per minute. However, she notices that the number of hits during the day-time (9.009.00 am to 9.009.00 pm) is usually about twice the number of hits during the night-time (9.009.00 pm to 9.009.00 am).

(b)

(i) Explain why this fact contradicts the owner’s belief that the number of hits per minute follows a Poisson distribution.

(1)

(ii) Specify separate Poisson distributions that might be suitable models for the number of hits during the day-time and during the night-time.

(2)
题目中文翻译

一名网站所有者发现，他的网站每分钟平均收到 0.30.3 次访问。他认为每分钟的访问次数服从泊松分布。

(a) 假设网站所有者的看法正确。

(i) 求一段 1010 分钟的时间内至少有 44 次访问的概率。

(3)

(ii) 用适当的近似分布求一段 33 小时的时间内访问次数少于 4040 的概率。

(4)

一名朋友同意这个网站平均每分钟收到 0.30.3 次访问。然而她注意到，白天（上午 99 点至晚上 99 点）的访问次数通常约为夜间（晚上 99 点至上午 99 点）的两倍。

(b)

(i) 解释为什么这一事实与网站所有者认为每分钟的访问次数服从泊松分布的看法相矛盾。

(1)

(ii) 分别写出一个适合描述白天访问次数的泊松分布和一个适合描述夜间访问次数的泊松分布。

(2)
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)(i)

解法一

思路

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10分钟的泊松参数为3，至少4次取不超过3次的补集。

答题过程

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Let H10H_{10} be the number of hits in 10 minutes. Then H10∼Po(10(0.3))=Po(3)H_{10}\sim\mathrm{Po}(10(0.3))=\mathrm{Po}(3).

P(H10≥4)= 1−e−3(1+3) −e−3(322!+333!)= 0.352768…= 0.353(3 s.f.).\begin{align*} \mathrm{P}(H_{10}\geq4)=&\,1-e^{-3}(1+3)\\[4mm] &\,-e^{-3}\bigg(\frac{3^2}{2!}+\frac{3^3}{3!}\bigg)\\[4mm] =&\,0.352768\ldots\\[4mm] =&\,\boxed{0.353}\quad\text{(3 s.f.).} \end{align*}

(a)(ii)

解法一

思路

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3小时为180分钟，参数为54，采用正态近似并将少于40次修正为39.5。

答题过程

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Over three hours, H180∼Po(180(0.3))=Po(54)H_{180}\sim\mathrm{Po}(180(0.3))=\mathrm{Po}(54). Using H180≈N(54,54)H_{180}\approx\mathrm{N}(54,54) with a continuity correction,

P(H180<40)≈ Φ(39.5−5454)= 0.0242364…= 0.0242(3 s.f.).\begin{align*} \mathrm{P}(H_{180}<40)\approx&\,\Phi\bigg(\frac{39.5-54}{\sqrt{54}}\bigg)\\[4mm] =&\,0.0242364\ldots\\[4mm] =&\,\boxed{0.0242}\quad\text{(3 s.f.).} \end{align*}

(b)(i)

解法一

思路

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白天和夜晚平均速率不同，与全天采用同一个固定参数的泊松模型所要求的恒定平均速率矛盾。

答题过程

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The average rate of hits is not constant throughout the day, so a single Poisson model with a constant mean of 0.3 hits per minute is inappropriate.

(b)(ii)

解法一

思路

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白天夜晚各12小时，平均每分钟速率为0.3且白天为夜晚两倍，先解两个速率，再说明模型对应的时长。

答题过程

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Let rr be the night-time mean rate per minute. The day-time rate is 2r2r. Since both periods last 12 hours,

r+2r2=0.3⟹r=0.2.\frac{r+2r}{2}=0.3 \quad\Longrightarrow\quad r=0.2.

Suitable models for the number of hits in one minute are

 Hday∼Po(0.4), Hnight∼Po(0.2).\begin{align*} &\,\boxed{H_{\mathrm{day}}\sim\mathrm{Po}(0.4)},\\[4mm] &\,\boxed{H_{\mathrm{night}}\sim\mathrm{Po}(0.2)}. \end{align*}

Equivalently, the totals over the full 12-hour periods may be modelled by Po(288)\mathrm{Po}(288) during the day-time and Po(144)\mathrm{Po}(144) during the night-time.