题目

Problem

The lengths of the leaves of a certain type of tree are normally distributed with mean $\mu\,\mathrm{cm}$ and standard deviation $\sigma\,\mathrm{cm}$ . It is known that 20% of the leaves have a length less than $3.60\,\mathrm{cm}$ and 70% of the leaves have a length greater than $5.40\,\mathrm{cm}$ .

(a) Find the value of $\mu$ and the value of $\sigma$ .

(5)

The lengths of 600 randomly chosen leaves of a different type of tree are normally distributed with mean $4.63\,\mathrm{cm}$ and standard deviation $0.55\,\mathrm{cm}$ .

(b) How many of these 600 leaves would you expect to have a length within $0.6\,\mathrm{cm}$ of the mean length?

(4)

题目中文翻译

某种树的叶片长度服从均值为 $\mu\,\mathrm{cm}$ 、标准差为 $\sigma\,\mathrm{cm}$ 的正态分布。已知 20% 的叶片长度小于 $3.60\,\mathrm{cm}$ ，70% 的叶片长度大于 $5.40\,\mathrm{cm}$ 。

(a) 求 $\mu$ 和 $\sigma$ 的值。

(5)

随机选取另一种树的 600 片叶子，其长度服从均值为 $4.63\,\mathrm{cm}$ 、标准差为 $0.55\,\mathrm{cm}$ 的正态分布。

(b) 预计这 600 片叶子中，有多少片的长度与平均长度之差在 $0.6\,\mathrm{cm}$ 以内？

(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

展开

先把两条概率都改成左尾累计概率，查对应的标准分数。两次标准化得到关于均值和标准差的线性方程，再相减消元。

答题过程

展开

Let $L\sim\mathrm{N}(\mu,\sigma^2)$ . The given probabilities imply

\mathrm{P}(L<3.60)=0.20,\qquad \mathrm{P}(L<5.40)=0.30.

The corresponding standard normal quantiles give

\begin{align*} \frac{3.60-\mu}{\sigma}=&\,-0.841621\ldots,\\[4mm] \frac{5.40-\mu}{\sigma}=&\,-0.524401\ldots. \end{align*}

Subtracting the first equation from the second,

\begin{align*} 1.80=&\,\bigl(0.841621\ldots-0.524401\ldots\bigr)\sigma,\\[4mm] \sigma=&\,5.67428\ldots. \end{align*}

Substituting into the first equation,

\mu=3.60+0.841621\ldots\,\sigma=8.375596\ldots.

Hence

\boxed{\mu\approx8.38,\qquad\sigma\approx5.67}.

(b)

解法一

思路

展开

“与均值相差不超过0.6”给出均值两侧对称的区间。先标准化求区间概率，再乘600得到预计叶片数。

答题过程

展开

For this type of leaf, $L\sim\mathrm{N}(4.63,0.55^2)$ . The required interval is $4.03<L<5.23$ , so

\begin{align*} \mathrm{P}(|L-4.63|<0.6) =&\,\mathrm{P}\biggl(-\frac{0.6}{0.55}<Z<\frac{0.6}{0.55}\biggr)\\[4mm] =&\,2\Phi(1.090909\ldots)-1\\[4mm] =&\,0.724687\ldots. \end{align*}

The expected number is $600(0.724687\ldots)=434.812\ldots$ , giving 435 leaves.

CIE 9709 2026 June Paper 52 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程