题目

Problem

The heights of the female students at Breven college are normally distributed:

$90\%$ of the female students have heights less than $182.7\,\mathrm{cm}$ .
$40\%$ of the female students have heights less than $162.5\,\mathrm{cm}$ .

(a) Find the mean and the standard deviation of the heights of the female students at Breven college.

(5)

Ten female students are chosen at random from those at Breven college.

(b) Find the probability that fewer than 8 of these 10 students have heights more than $162.5\,\mathrm{cm}$ .

(3)

题目中文翻译

Breven college 女学生的身高服从正态分布：

$90\%$ 的女学生身高小于 $182.7\,\mathrm{cm}$ 。
$40\%$ 的女学生身高小于 $162.5\,\mathrm{cm}$ 。

(a) 求 Breven college 女学生身高的均值和标准差。

(5)

从 Breven college 的女学生中随机选取 10 人。

(b) 求这 10 人中身高大于 $162.5\,\mathrm{cm}$ 的少于 8 人的概率。

(3)

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

利用两个左尾概率求标准分数，列出含均值和标准差的两个线性方程。0.4小于0.5，对应负标准分数。

答题过程

展开

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

\frac{182.7-\mu}{\sigma}=\Phi^{-1}(0.9)=1.281551\ldots,

\frac{162.5-\mu}{\sigma}=\Phi^{-1}(0.4)=-0.253347\ldots.

Subtracting gives

\begin{align*} 20.2=&\,(1.281551\ldots+0.253347\ldots)\sigma,\\[4mm] \sigma=&\,13.160477\ldots. \end{align*}

Substituting into the first equation,

\begin{align*} \mu=&\,182.7-(1.281551\ldots)(13.160477\ldots)\\[4mm] =&\,165.834168\ldots. \end{align*}

Hence the mean is $165.8\,\mathrm{cm}$ and the standard deviation is $13.2\,\mathrm{cm}$ .

(b)

解法一

思路

展开

大于162.5的概率为0.6。少于8的补集只需计算8、9、10三项。

答题过程

展开

For each student, $\mathrm{P}(H>162.5)=1-0.4=0.6$ . Let $X\sim\mathrm{B}(10,0.6)$ count students taller than 162.5 cm. Then

\begin{align*} \mathrm{P}(X<8) =&\,1-\bigl[\binom{10}{8}(0.6)^8(0.4)^2\\[4mm] &\,\hspace{2pt}+10(0.6)^9(0.4)+(0.6)^{10}\bigr]\\[4mm] =&\,0.8327102464\\[4mm] \approx&\,\boxed{0.833}. \end{align*}

CIE 9709 2024 November Paper 51 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程