题目

Problem

Flogcar is a car hire company that only allows cars to be hired for single days. The numbers of cars hired on different days are statistically independent. The number of cars hired on a randomly selected day from Flogcar is modelled by a Poisson distribution with mean 4

(a) Calculate the probability that exactly 6 cars are hired from Flogcar on a randomly selected day.

(2)

(b) Calculate the probability that on 2 randomly selected days, the total number of cars hired from Flogcar is between 3 and 7 inclusive.

(3)

(c) Use a suitable approximation to find the probability that on 7 randomly selected days, the total number of cars hired from Flogcar is more than 30

(5)

The probability that a car hired from Flogcar is returned in good condition is 0.97

In a randomly chosen period, Flogcar hires out 100 cars.

(d) (i) Explain why the distribution of the number of cars returned in good condition would not be well approximated by the Poisson distribution.

(ii) By considering the number of cars hired out that are not returned in good condition, use a Poisson approximation to estimate the probability that more than 95 of the cars hired out in this period are returned in good condition. Show your working clearly.

(1)

(3)

(Total for Question 1 is 14 marks)

题目中文翻译

Flogcar 是一家租车公司，只允许按天租车。不同日子租出的汽车数量彼此独立。随机选取一天，Flogcar 租出的汽车数量服从均值为 4 的泊松分布。

(a) 计算随机选取的一天恰好租出 6 辆车的概率。

(b) 计算随机选取 2 天，Flogcar 租出汽车总数在 3 到 7 辆之间（含端点）的概率。

一辆从 Flogcar 租出的汽车被完好归还的概率为 0.97。

在某个随机选取的时期内，Flogcar 共租出 100 辆车。

(d) (i) 解释为什么“完好归还的汽车数量”分布不适合用泊松分布近似。

(ii) 通过考虑“未完好归还的汽车数量”，使用泊松近似估计该时期内完好归还的汽车超过 95 辆的概率，并清楚写出你的过程。

（第 1 题共 14 分）

解答

(a)

解法一

思路

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一天租出的汽车数量服从均值为 4 的泊松分布，直接代入泊松分布的概率公式求恰好 6 辆。

答题过程

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Let $H$ be the number of cars hired on one day. Then

H\sim\operatorname{Po}(4).

Therefore,

\begin{align*} P(H=6) =&\,e^{-4}\frac{4^6}{6!} \\[4mm] =&\,0.104195\ldots \\[4mm] =&\,\boxed{0.104}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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独立泊松随机变量之和仍服从泊松分布，均值等于各均值之和。因此两天总数的均值为 8；“3 到 7 辆（含端点）”可用两个累积概率相减。

答题过程

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Let $J$ be the total number of cars hired on two days. Since the daily numbers are independent,

J\sim\operatorname{Po}(8).

Hence,

\begin{align*} P(3\leqslant J\leqslant7) =&\,P(J\leqslant7)-P(J\leqslant2) \\[4mm] =&\,0.45296\ldots-0.01375\ldots \\[4mm] =&\,0.43921\ldots \\[4mm] =&\,\boxed{0.439}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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七天总数服从均值为 28 的泊松分布。均值足够大，可用均值和方差均为 28 的正态分布近似。事件“超过 30”即至少为 31，所以连续性修正后的边界是 30.5。

答题过程

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Let $K$ be the total number of cars hired on seven days. Exactly,

K\sim\operatorname{Po}(28).

Using a normal approximation,

K\approx N(28,28).

With a continuity correction,

\begin{align*} P(K>30) \approx&\,P(Y>30.5) \\[4mm] =&\,P\left( Z>\frac{30.5-28}{\sqrt{28}} \right) \\[4mm] =&\,P(Z>0.47246\ldots) \\[4mm] =&\,\boxed{0.318}\quad\text{(3 s.f.)}. \end{align*}

Using normal tables with $z=0.47$ gives $0.319$ , which is also accepted by the official mark scheme.

(d)(i)

解法一

思路

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用泊松分布近似二项分布时，成功概率应较小；这里“完好归还”的概率为 0.97，显然不小。也可看到二项分布的均值与方差相差很大，不符合泊松分布均值等于方差的特征。

答题过程

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The probability of a car being returned in good condition is $0.97$ , which is not small. Therefore, the binomial distribution for the number returned in good condition is not well approximated by a Poisson distribution.

(d)(ii)

解法一

思路

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改为统计“未完好归还”的汽车后，单辆车对应的概率是 0.03，满足小概率条件。其二项分布可用均值 $100\times0.03=3$ 的泊松分布近似。超过 95 辆完好归还等价于至多 4 辆未完好归还。

答题过程

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Let $L$ be the number of cars not returned in good condition. Exactly,

L\sim\operatorname{B}(100,0.03).

Using a Poisson approximation,

L\approx\operatorname{Po}(3).

More than 95 cars returned in good condition means at most 4 cars were not returned in good condition. Therefore, the required probability is approximated by

\begin{align*} P(L\leqslant4) =&\,0.815263\ldots \\[4mm] =&\,\boxed{0.815}\quad\text{(3 s.f.)}. \end{align*}

IAL 2020 Jan S2 Q1

题目

解答

(a)

解法一

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答题过程

(b)

解法一

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(c)

解法一

思路

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(d)(i)

解法一

思路

答题过程

(d)(ii)

解法一

思路

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