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CIE 9709 2025 June Paper 61 Q4

A Level / CIE / S2

CIE 9709 2025 June Paper 61 Paper · Question 4

题目

Problem

At an entertainment centre, the cost for using a particular video game is $0.40 per minute. The number of minutes for which people use the video game has mean 1515 and variance 99.

(a) Find the mean and variance of the amount people pay for using the video game.

(3)

Each day, 3535 people independently use the video game.

(b) Find the mean and variance of the total amount paid by 3535 people.

(3)
题目中文翻译

在一家娱乐中心,玩某款电子游戏的费用为每分钟 0.400.40 美元。人们玩这款游戏的时长以分钟为单位,均值为 1515,方差为 99。

(a) 求人们玩这款游戏所支付金额的均值和方差。

(3)

每天有 3535 人各自独立地玩这款游戏。

(b) 求 3535 人支付的总金额的均值和方差。

(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

展开

支付金额是使用时间乘以 0.400.40;均值按一次系数缩放,方差按系数平方缩放。

答题过程

展开

Let TT be a person’s playing time and C=0.40TC=0.40T the amount paid, in dollars. Then

E(C)=0.40(15)=6,\boxed{\mathrm E(C)=0.40(15)=6},

and

Var⁡(C)=0.402(9)=1.44.\boxed{\operatorname{Var}(C)=0.40^2(9)=1.44}.

(b)

解法一

思路

展开

三十五位顾客是独立使用时间的总和,方差相加,不能把一个人的费用乘以三十五后平方。

答题过程

展开

Let A=C1+⋯+C35A=C_1+\cdots+C_{35} be the total amount paid. Using independence and part (a),

E(A)=35(6)=210,\boxed{\mathrm E(A)=35(6)=210},

and

Var⁡(A)=35(1.44)=50.4.\boxed{\operatorname{Var}(A)=35(1.44)=50.4}.