题目

Problem

The data produced by a certain data entry firm always include a small number of incorrect characters that occur at random. The proportion of incorrect characters is denoted by $p$ , and experience has shown that $p=0.0001$ . A particular data set from the firm contains $14\,500$ characters, of which $X$ characters are incorrect.

(a) Use a suitable approximating distribution to find $\mathrm{P}(X<4)$ .

(3)

The firm’s management wishes to decrease the value of $p$ by giving their employees some training. Their aim is that, for a data set containing $14\,500$ characters, the value of $\mathrm{P}(X=0)$ for the new value of $p$ should be double the value of $\mathrm{P}(X=0)$ when $p=0.0001$ .

(b) Use a suitable approximating distribution to find the new value of $p$ .

(3)

题目中文翻译

某数据录入公司所生成的数据总会包含少量随机出现的错字。用 $p$ 表示错字所占的比例。经验表明 $p=0.0001$ 。该公司的一份数据包含 $14\,500$ 个字符，其中有 $X$ 个错字。

(a) 用一个合适的近似分布求 $\mathrm{P}(X<4)$ 。

(3)

公司管理层希望通过培训员工来降低 $p$ 。他们的目标是，对于包含 $14\,500$ 个字符的数据，在新的 $p$ 值下， $\mathrm{P}(X=0)$ 应为 $p=0.0001$ 时 $\mathrm{P}(X=0)$ 的两倍。

(b) 用一个合适的近似分布求新的 $p$ 值。

(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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字符数大、错误比例小，使用二项分布的 Poisson 近似；小于四个错误包括 $0$ 到 $3$ 。

答题过程

展开

With a large number of characters and small error probability, use

X\approx\mathrm{Po}(14500(0.0001))=\mathrm{Po}(1.45).

Then

\begin{align*} \mathrm P(X<4) \approx&\,\mathrm e^{-1.45}(1+1.45)\\[4mm] &\,+\mathrm e^{-1.45}\bigg(\frac{1.45^2}{2!}+\frac{1.45^3}{3!}\bigg)\\[4mm] \approx&\,\boxed{0.940}. \end{align*}

(b)

解法一

思路

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错误比例改变后，Poisson 参数改成 $14500p$ ；“零错误概率加倍”应把倍数放在旧概率那一侧，再取对数。

答题过程

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For the new proportion $p$ , use $X\approx\mathrm{Po}(14500p)$ . The required change in the zero-error probability gives

\begin{align*} \mathrm e^{-14500p}=&\,2\mathrm e^{-1.45},\\[4mm] -14500p=&\,\ln2-1.45,\\[4mm] p=&\,\frac{1.45-\ln2}{14500}. \end{align*}

Thus $\boxed{p\approx0.0000522}$ . This is positive and smaller than $0.0001$ , as required.

CIE 9709 2025 November Paper 61 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