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CIE 9709 2024 November Paper 51 Q5

A Level / CIE / S1

CIE 9709 2024 Nov Paper 51 Paper · Question 5

题目

Problem

The weights of the green apples sold by a shop are normally distributed with mean 90 grams and standard deviation 8 grams.

(a) Find the probability that a randomly chosen green apple weighs between 83 grams and 95 grams.

(4)

(b) The shop also sells red apples. 60%60\% of the red apples sold by the shop weigh more than 80 grams. 160 red apples are chosen at random from the shop.

Use a suitable approximation to find the probability that fewer than 105 of the chosen red apples weigh more than 80 grams.

(5)
题目中文翻译

某商店出售的青苹果的重量服从正态分布，均值为 90 克，标准差为 8 克。

(a) 随机选取一个青苹果，求其重量在 83 克与 95 克之间的概率。

(4)

(b) 这家商店也出售红苹果。该店出售的红苹果中，60%60\% 的重量超过 80 克。从店中随机选取 160 个红苹果。

使用适当的近似方法，求所选红苹果中重量超过 80 克的少于 105 个的概率。

(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

展开

分别标准化上下界，求两个累计概率之差。

答题过程

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Let W∼N(90,82)W\sim\mathrm{N}(90,8^2) be the weight of a green apple in grams. Then

P(83<W<95)= Φ(95−908)−Φ(83−908)= Φ(0.625)−Φ(−0.875)= 0.543227…≈ 0.543.\begin{align*} \mathrm{P}(83<W<95) =&\,\Phi\biggl(\frac{95-90}{8}\biggr) -\Phi\biggl(\frac{83-90}{8}\biggr)\\[4mm] =&\,\Phi(0.625)-\Phi(-0.875)\\[4mm] =&\,0.543227\ldots\\[4mm] \approx&\,\boxed{0.543}. \end{align*}

(b)

解法一

思路

展开

人数服从二项分布，用正态近似。少于105即至多104，连续性修正为104.5。

答题过程

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Let X∼B(160,0.6)X\sim\mathrm{B}(160,0.6) count red apples weighing more than 80 grams. Since np=96np=96 and n(1−p)=64n(1-p)=64 are both greater than 5, use

Y∼N(96,38.4),38.4=160(0.6)(0.4).Y\sim\mathrm{N}(96,38.4),\qquad 38.4=160(0.6)(0.4).

With a continuity correction,

P(X<105)= P(X≤104)≈ P(Y<104.5)= Φ(104.5−9638.4)= 0.914918…≈ 0.915.\begin{align*} \mathrm{P}(X<105) =&\,\mathrm{P}(X\leq104)\\[4mm] \approx&\,\mathrm{P}(Y<104.5)\\[4mm] =&\,\Phi\biggl(\frac{104.5-96}{\sqrt{38.4}}\biggr)\\[4mm] =&\,0.914918\ldots\\[4mm] \approx&\,\boxed{0.915}. \end{align*}