题目

Problem

A company sells small and large bags of rice. The masses of the small bags of rice are normally distributed with mean $1.20\,\mathrm{kg}$ and standard deviation $0.16\,\mathrm{kg}$ .

(a) In a random sample of 500 of these small bags of rice, how many would you expect to have a mass greater than $1.26\,\mathrm{kg}$ ?

(4)

The masses of the large bags of rice are normally distributed with mean $2.50\,\mathrm{kg}$ and standard deviation $\sigma\,\mathrm{kg}$ . $20\%$ of these large bags of rice have a mass less than $2.40\,\mathrm{kg}$ .

(b) Find the value of $\sigma$ .

(3)

A random sample of 80 large bags of rice is chosen.

(c) Use a suitable approximation to find the probability that fewer than 22 of these large bags of rice have a mass less than $2.40\,\mathrm{kg}$ .

(5)

题目中文翻译

一家公司销售小袋和大袋大米。小袋大米的质量服从均值为 $1.20\,\mathrm{kg}$ 、标准差为 $0.16\,\mathrm{kg}$ 的正态分布。

(a) 在随机抽取的 500 袋这种小袋大米中，预计有多少袋的质量大于 $1.26\,\mathrm{kg}$ ？

(4)

大袋大米的质量服从均值为 $2.50\,\mathrm{kg}$ 、标准差为 $\sigma\,\mathrm{kg}$ 的正态分布。这些大袋大米中有 $20\%$ 的质量小于 $2.40\,\mathrm{kg}$ 。

(b) 求 $\sigma$ 的值。

(3)

随机抽取 80 袋大袋大米。

(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

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先求单袋超过1.26的概率，保留精度乘500，最后取整数袋数。

答题过程

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Let $M\sim\mathrm{N}(1.20,0.16^2)$ . Then

\begin{align*} \mathrm{P}(M>1.26) =&\,1-\Phi\biggl(\frac{1.26-1.20}{0.16}\biggr)\\[4mm] =&\,1-\Phi(0.375)=0.353830233\ldots. \end{align*}

The expected number is

500(0.353830233\ldots)=176.915116\ldots.

Thus we would expect 177 bags.

(b)

解法一

思路

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小于2.40的概率为左尾0.20，用负分位点求标准差。

答题过程

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\frac{2.40-2.50}{\sigma}=\Phi^{-1}(0.20)=-0.841621\ldots.

Hence

\sigma=\frac{-0.10}{-0.841621\ldots} =0.118818294\ldots\approx\boxed{0.119}.

(c)

解法一

思路

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袋数使用题给0.20为成功概率，不需重新用(b)的舍入标准差计算。少于22修正为21.5。

答题过程

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Let $X\sim\mathrm{B}(80,0.20)$ count bags weighing less than 2.40 kg. Since $np=16$ and $n(1-p)=64$ are both greater than 5, use $Y\sim\mathrm{N}(16,12.8)$ .

With a continuity correction,

\begin{align*} \mathrm{P}(X<22) =&\,\mathrm{P}(X\leq21)\\[4mm] \approx&\,\mathrm{P}(Y<21.5)\\[4mm] =&\,\Phi\biggl(\frac{21.5-16}{\sqrt{12.8}}\biggr)\\[4mm] =&\,0.937889669\ldots\\[4mm] \approx&\,\boxed{0.938}. \end{align*}

CIE 9709 2024 March Paper 52 Q4

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