题目
The random variable has the distribution . A random sample of values of is chosen, and the sample mean, , is found.
(a) Find . You are not expected to use a continuity correction.
(b) State why the Central Limit Theorem was needed in the calculation in part (a).
题目中文翻译
随机变量 服从分布 。随机选取 的 个取值组成样本,并求出样本均值 。
(a) 求 。不要求使用连续性修正。
(b) 说明为什么在 (a) 的计算中需要使用中心极限定理。
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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先求单个二项变量的均值与方差,再转为样本均值的均值和方差;按题意不做连续性修正。
答题过程
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For ,
By the Central Limit Theorem, for a random sample of size ,
Without a continuity correction, as instructed,
(b)
解法一
思路
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原变量是二项分布而非正态;用中心极限定理来支持样本均值的正态近似。
答题过程
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The population distribution of is binomial, not normal. The Central Limit Theorem is needed to justify approximating the sample-mean distribution by a normal distribution for the large sample.