题目

Problem

The weights of oranges can be modelled by a normal distribution with mean 131 grams and standard deviation 54 grams. Oranges are classified as small, medium or large. A large orange weighs at least 184 grams and $20\%$ of oranges are classified as small.

(a) Find the percentage of oranges that are classified as large.

(3)

(b) Find the greatest possible weight of a small orange.

(3)

题目中文翻译

橙子的重量可用均值为 131 克、标准差为 54 克的正态分布建模。橙子分为小、中、大三类。大橙子的重量至少为 184 克，且有 $20\%$ 的橙子被归为小橙子。

(a) 求被归为大橙子的橙子所占的百分比。

(3)

(b) 求小橙子的最大可能重量。

(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

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正态模型连续，至少184与大于184概率相同。求右尾后乘100变成百分比。

答题过程

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Let $W\sim\mathrm{N}(131,54^2)$ . Then

\begin{align*} \mathrm{P}(W\geq184) =&\,1-\Phi\biggl(\frac{184-131}{54}\biggr)\\[4mm] =&\,0.163177681\ldots. \end{align*}

Thus $16.3\%$ of oranges are classified as large.

(b)

解法一

思路

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小橙子上界对应左尾0.20，用负分位点求重量。

答题过程

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Let $w$ be the upper weight boundary for small oranges. Then

\frac{w-131}{54}=\Phi^{-1}(0.20)=-0.841621\ldots.

Therefore

w=131+54(-0.841621\ldots) =85.552453\ldots\approx\boxed{85.6\text{ grams}}.

CIE 9709 2024 June Paper 52 Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程