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CIE 9709 2022 November Paper 51 Q1

A Level / CIE / S1

CIE 9709 2022 Nov Paper 51 Paper · Question 1

题目

Problem

The probability distribution table for a random variable XX is shown below.

xx−2-2−1-10.50.51122
P(X=x)\mathrm{P}(X=x)0.120.12ppqq0.160.160.30.3

Given that E(X)=0.28\mathrm{E}(X)=0.28, find the value of pp and the value of qq.

(4)

题目中文翻译

随机变量 XX 的概率分布表如下。

xx−2-2−1-10.50.51122
P(X=x)\mathrm{P}(X=x)0.120.12ppqq0.160.160.30.3

已知 E(X)=0.28\mathrm{E}(X)=0.28,求 pp 和 qq 的值。

(4)

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

解法一

思路

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概率总和和给定均值列两个方程,消元求p、q。

答题过程

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Since the probabilities sum to 1,

p+q=1−0.12−0.16−0.30=0.42.p+q=1-0.12-0.16-0.30=0.42.

Using E(X)=0.28\mathrm{E}(X)=0.28,

−2(0.12)−p+0.5q+0.16+2(0.30)=0.28,-2(0.12)-p+0.5q+0.16+2(0.30)=0.28,

so

−p+0.5q=−0.24.-p+0.5q=-0.24.

Adding the two equations gives 1.5q=0.181.5q=0.18, so q=0.12q=0.12. Then p=0.42−0.12=0.30p=0.42-0.12=0.30. Thus

p=0.30,q=0.12.\boxed{p=0.30,\qquad q=0.12}.