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。。 · 2025年07月09日

这道题不是考假设检验

NO.PZ2023091601000058

问题如下:

A risk manager is examining aHong Kong trader’s profit and loss record for the last week, as shown in thetable below:

The profits andlosses are normally distributed with a mean of 4.5 million HKD and assume thattransaction costs can be ignored. Part of the t-table is provided below:

Accordingto the information provided above, what is the probability that this traderwill record a profit of at least HKD 30 million on the first trading day ofnext week?

选项:

A.

About 15%

B.

About 20%

C.

About 80%

D.

About 85%

解释:

When the populationmean and population variance are not known, the t-statistic can be used toanalyze the distribution of the sample mean.

Sample mean = (10 +80 + 90 - 60 + 30)/5 = 30

Unbiased samplevariance = (1/4)[ (-20)^2 + 50^2 + 60^2 + (-90)^2 + 0^2 ] = 14600/4 = 3650

Unbiased samplestandard deviation = 60.4152

Sample standard error= (sample standard deviation)/√5 = 27.0185

Population mean ofreturn distribution = 4.5 (million HKD)

Therefore thet-statistic = (30 – population mean)/Sample standard error = (30 - 4.5)/27.02 =0.9438.

Because we are usingthe sample mean in the analysis, we must remove 1 degree of freedom beforeconsulting the t-table; therefore 4 degrees of freedom are used. According tothe table, the closest possibility is 0.2 = 20%.

​一般关于这类题目下意识以为是假设检验,实际上是检验profit不小于30的概率,进行标准化后查分布表

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NO.PZ2023091601000058问题如下 A risk manager is examining aHong Kong trar’s profit anloss recorfor the last week, shown in thetable below:The profits anosses are normally stributewith a meof 4.5 million HKanassume thattransaction costs cignore Part of the t-table is provibelow: Accorngto the information proviabove, whis the probability ththis trarwill recora profit of least HK30 million on the first trang y ofnext week? A.About 15% B.About 20% C.About 80% About 85% When the populationmeanpopulation varianare not known, the t-statistic cusetoanalyze the stribution of the sample mean. Sample me= (10 +80 + 90 - 60 + 30)/5 = 30 Unbiasesamplevarian= (1/4)[ (-20)^2 + 50^2 + 60^2 + (-90)^2 + 0^2 ] = 14600/4 = 3650 Unbiasesamplestanrviation = 60.4152 Sample stanrerror= (sample stanrviation)/√5 = 27.0185 Population meofreturn stribution = 4.5 (million HK Therefore thet-statistic = (30 – population mean)/Sample stanrerror = (30 - 4.5)/27.02 =0.9438. Because we are usingthe sample mein the analysis, we must remove 1 gree of freem beforeconsulting the t-table; therefore 4 grees of freem are use Accorng tothe table, the closest possibility is 0.2 = 20%. 1、这道题就是求P(X=30million)吧?而不是P(X大于等于30million)?2、为什么30减去的是总体均值,而不是样本均值?

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