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IAL 2020 Oct S3 Q6

A Level / Edexcel / S3

IAL 2020 Oct Paper · Question 6

题目

Problem

The number of toasters sold by a shop each week may be modelled by a Poisson distribution with mean 4.

A random sample of 35 weeks is taken and the mean number of toasters sold per week is found.

(a) Write down the approximate distribution for the mean number of toasters sold per week from a random sample of 35 weeks.

(2)

The number of kettles sold by the shop each week may be modelled by a Poisson distribution with mean λ\lambda.

A random sample of 40 weeks is taken and the mean number of kettles sold per week is found. The width of the 99% confidence interval for λ\lambda is 2.6.

(b) Find an estimate for λ\lambda.

(4)

A second, independent random sample of 40 weeks is taken and a second 99% confidence interval for λ\lambda is found.

(c) Find the probability that only one of these two confidence intervals contains λ\lambda.

(2)
(Total 8 marks)
题目中文翻译

一家商店每周售出的烤面包机数量可以用均值为 4 的泊松分布来建模。

抽取 35 周的随机样本，并求每周售出烤面包机数量的样本均值。

(a) 写出从 35 周随机样本得到的每周售出烤面包机数量均值的近似分布。

该商店每周售出的水壶数量可以用均值为 λ\lambda 的泊松分布来建模。

抽取 40 周的随机样本，并求每周售出水壶数量的样本均值。λ\lambda 的 99% 置信区间宽度为 2.6。

(b) 求 λ\lambda 的一个估计值。

再抽取第二个彼此独立的 40 周随机样本，并得到 λ\lambda 的第二个 99% 置信区间。

(c) 求恰好只有这两个置信区间中的一个包含 λ\lambda 的概率。

解答

(a)

解法一

思路

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单周销量服从均值和方差都为 44 的泊松分布。样本量 3535 较大，由中心极限定理，样本均值近似服从正态分布；其均值仍为 44，方差则是单个观测方差除以 3535。

答题过程

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For one week, the number sold has mean and variance 4. Therefore, by the central limit theorem,

Tˉ∼˙N(4,435).\boxed{\bar T\mathrel{\dot\sim}N\bigg(4,\frac{4}{35}\bigg)}.

(b)

解法一

思路

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每周水壶销量服从泊松分布，因此方差也为 λ\lambda。40 周样本均值近似服从 N(λ,λ/40)N(\lambda,\lambda/40)。99%99\% 双侧置信区间的宽度等于两倍误差界，即 2z0.995λ/402z_{0.995}\sqrt{\lambda/40}；令它等于 2.62.6 后解出 λ\lambda。

答题过程

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Let Kˉ\bar K be the sample mean number of kettles sold. Then

Kˉ∼˙N(λ,λ40).\bar K\mathrel{\dot\sim}N\bigg(\lambda,\frac{\lambda}{40}\bigg).

For a 99%99\% confidence interval,

z0.995=2.5758.z_{0.995}=2.5758.

The width of the confidence interval is therefore

2(2.5758)λ40=2.6.2(2.5758)\sqrt{\frac{\lambda}{40}}=2.6.

Hence

λ40= 0.50470…λ40= 0.25472…λ= 10.1888…\begin{align*} \sqrt{\frac{\lambda}{40}}=&\,0.50470\ldots\\ \frac{\lambda}{40}=&\,0.25472\ldots\\ \lambda=&\,10.1888\ldots \end{align*}

Thus, to 3 significant figures,

λ^=10.2.\boxed{\hat\lambda=10.2}.

(c)

解法一

思路

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每个 99%99\% 置信区间包含 λ\lambda 的概率为 0.990.99，不包含的概率为 0.010.01。两个样本相互独立，恰好一个区间包含 λ\lambda 有两种对称情况，因此把其中一种情况的概率乘以 22。

答题过程

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The two independent possibilities are:

  • the first interval contains λ\lambda and the second does not;
  • the first interval does not contain λ\lambda and the second does.

Therefore, the required probability is

P(only one contains λ)= 2(0.99)(0.01)= 0.0198.\begin{align*} P(\text{only one contains }\lambda)=&\,2(0.99)(0.01)\\ =&\,\boxed{0.0198}. \end{align*}