题目

Problem

The lengths of Western bluebirds are normally distributed with mean $16.5\,\mathrm{cm}$ and standard deviation $0.6\,\mathrm{cm}$ .

A random sample of 150 of these birds is selected.

(a) How many of these 150 birds would you expect to have length between $15.4\,\mathrm{cm}$ and $16.8\,\mathrm{cm}$ ?

(4)

The lengths of Eastern bluebirds are normally distributed with mean $18.4\,\mathrm{cm}$ and standard deviation $\sigma\,\mathrm{cm}$ . It is known that 72% of Eastern bluebirds have length greater than $17.1\,\mathrm{cm}$ .

(b) Find the value of $\sigma$ .

(3)

A random sample of 120 Eastern bluebirds is chosen.

(c) Use an approximation to find the probability that fewer than 80 of these 120 bluebirds have length greater than $17.1\,\mathrm{cm}$ .

(5)

题目中文翻译

西方蓝鸲的体长服从正态分布，均值为 $16.5\,\mathrm{cm}$ ，标准差为 $0.6\,\mathrm{cm}$ 。

随机选取 150 只这种鸟作为样本。

(a) 预计这 150 只鸟中有多少只的体长在 $15.4\,\mathrm{cm}$ 与 $16.8\,\mathrm{cm}$ 之间？

(4)

东方蓝鸲的体长服从正态分布，均值为 $18.4\,\mathrm{cm}$ ，标准差为 $\sigma\,\mathrm{cm}$ 。已知 72% 的东方蓝鸲体长大于 $17.1\,\mathrm{cm}$ 。

(b) 求 $\sigma$ 的值。

(3)

随机选取 120 只东方蓝鸲作为样本。

(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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求两个边界之间的正态概率，再乘150，最后给出整数鸟数。

答题过程

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Let $L\sim\mathrm{N}(16.5,0.6^2)$ . Then

\begin{align*} \mathrm{P}(15.4<L<16.8) =&\,\Phi\biggl(\frac{16.8-16.5}{0.6}\biggr) -\Phi\biggl(\frac{15.4-16.5}{0.6}\biggr)\\[4mm] =&\,0.658085953\ldots. \end{align*}

The expected number is

150(0.658085953\ldots)=98.712893\ldots.

Thus we would expect 99 birds.

(b)

解法一

思路

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右尾0.72换为左尾0.28，对应负分位点。

答题过程

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\frac{17.1-18.4}{\sigma}=\Phi^{-1}(0.28)=-0.582841507\ldots.

Therefore

\sigma=\frac{-1.3}{-0.582841507\ldots} =2.230451990\ldots\approx\boxed{2.23}.

(c)

解法一

思路

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直接用题给大于17.1的概率0.72建二项分布，少于80修正为79.5。

答题过程

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Let $X\sim\mathrm{B}(120,0.72)$ . Since $np=86.4$ and $n(1-p)=33.6$ are both greater than 5, use $Y\sim\mathrm{N}(86.4,24.192)$ .

With a continuity correction,

\begin{align*} \mathrm{P}(X<80) =&\,\mathrm{P}(X\leq79)\\[4mm] \approx&\,\mathrm{P}(Y<79.5)\\[4mm] =&\,\Phi\biggl(\frac{79.5-86.4}{\sqrt{24.192}}\biggr)\\[4mm] =&\,0.080329839\ldots\approx\boxed{0.0803}. \end{align*}

CIE 9709 2023 June Paper 52 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程