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Lich · 2024年11月02日

如题

NO.PZ2023091601000058

问题如下:

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

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

According to the information provided above, what is the probability that this trader will record a profit of at least HKD 30 million on the first trading day of next week?

选项:

A.

About 15%

B.

About 20%

C.

About 80%

D.

About 85%

解释:

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

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

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

Unbiased sample standard deviation = 60.4152

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

Population mean of return distribution = 4.5 (million HKD)

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

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

有个小问题,这个题目给的t分布查表是不是单尾的呢?



1 个答案
已采纳答案

李坏_品职助教 · 2024年11月02日

嗨,努力学习的PZer你好:


对,是单尾的t分布表。


题目最后让你检测的命题是“profit of at least HKD 30 million”,也就是profit是否大于等于30 million。只要涉及到大于或小于的这种命题,就是单尾检验。

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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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