题目

Problem

The lengths, in cm, of the leaves of a particular type are modelled by the distribution $\mathrm{N}(5.2,\,1.5^2)$ .

(a) Find the probability that a randomly chosen leaf of this type has length less than $6\,\mathrm{cm}$ .

(2)

The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than $3\,\mathrm{cm}$ long and 95 are more than $8\,\mathrm{cm}$ long.

(b) Find estimates for the mean and standard deviation of the lengths of leaves of this type.

(5)

(c) In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean?

(4)

题目中文翻译

某种叶子的长度以厘米为单位，用分布 $\mathrm{N}(5.2,\,1.5^2)$ 建模。

(a) 求随机选取的一片这种叶子的长度小于 $6\,\mathrm{cm}$ 的概率。

(2)

另一种叶子的长度也用正态分布建模。一位科学家测量了这种叶子的一个含 500 片叶子的随机样本的长度，发现其中 46 片的长度小于 $3\,\mathrm{cm}$ ，95 片的长度大于 $8\,\mathrm{cm}$ 。

(b) 求这种叶子的长度的平均数和标准差的估计值。

(5)

(4)

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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标准化6cm并求左尾概率。

答题过程

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For $L\sim\mathrm{N}(5.2,1.5^2)$ ,

\mathrm{P}(L<6)=\Phi\biggl(\frac{6-5.2}{1.5}\biggr) =0.703098571\ldots\approx\boxed{0.703}.

(b)

解法一

思路

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样本比例46/500及95/500估计两尾概率，第二个换成左尾405/500，列两个分位点方程。

答题过程

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The estimated left-tail probabilities are

\begin{align*} \mathrm{P}(L<3)=&\,\frac{46}{500}=0.092,\\[4mm] \mathrm{P}(L<8)=&\,1-\frac{95}{500}=0.81. \end{align*}

Hence

\frac{3-\mu}{\sigma}=\Phi^{-1}(0.092)=-1.328539328\ldots,

\frac{8-\mu}{\sigma}=\Phi^{-1}(0.81)=0.877896295\ldots.

Subtracting gives

\sigma=\frac5{1.328539328\ldots+0.877896295\ldots} =2.266098292\ldots.

Then

\mu=3+1.328539328\ldots(2.266098292\ldots) =6.010600704\ldots.

The estimates are mean $6.01\,\mathrm{cm}$ and standard deviation $2.27\,\mathrm{cm}$ .

(c)

解法一

思路

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超过一个标准差是两尾之和，不能取中央区间。先算概率再乘2000。

答题过程

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\begin{align*} p=&\,\mathrm{P}(Z<-1)+\mathrm{P}(Z>1)\\[4mm] =&\,2\bigl[1-\Phi(1)\bigr]=0.317310507\ldots. \end{align*}

The expected number is

2000p=634.621015\ldots.

Thus the scientist would expect 635 leaves.

CIE 9709 2022 June Paper 51 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