概率分布作业

/*导入数据*/

import excel "C:\Users\86138\Desktop\20231213-发布作业-20231223 1540前提交\DowMarketCap.xlsx", sheet("DATA") firstrow clear

/*查看数据*/

describe

/*描述性统计*/

summarize

/*详细描述*/

summarize, detail

/*画出正态概率图*/

pnorm MarketCap ($billions)

/*画直方图*/

histogram MarketCap ($billions)

瑞典咖啡

a. What is the probability that someone in Sweden consumed more than 10 pounds of coffee in 2008?

解：

Z=(X-u)/rou=(10-18.04)/5=-1.608

查表得出Z= -1.608对应的百分比为5.37%，此概率为消费小于等于10磅咖啡的概率，因此消费大于10磅咖啡的概率为100%-5.37%=94.63%

  

b. What is the probability that someone in Sweden consumed between 3 and 5 pounds of coffee in 2008?

解：

Z=(X-u)/rou=

查表得出Z= -3.008对应的百分比为0.13%，Z= -2.608对应的百分比为0.45%   

因此消费咖啡量在3<X<5磅之间的百分比为0.45% -0.13%= 0.32%

  

c. What is the probability that someone in Sweden consumed less than 5 pounds of coffee in 2008?

解：

Z=(X-u)/rou=

查表得出Z= -2.608对应的百分比为0.45%，因此消费咖啡小于5磅的概率为0.45%。

  

d. 99% of the people in Sweden consumed less than how many pounds of coffee?

答：因为假设瑞典人均咖啡消费量近似分布为正态随机变量，概率99%对应的Z值查表为Z=2.33

根据公式Z=(X-u)/rou变形得出X=Zrou+u，所以X=2.33*5+18.04=29.69磅

因此，99%的瑞典人喝的咖啡少于29.69磅。

区间估计

a. Construct a 95% confidence interval estimate for the population mean difference between the actual length of the steel part and the specified length of the steel part.

答：首先导入数据，使用命令summarize得出样本均值X-bar=-0.00023，标准方差S=0.0016958

已知n=100，可求df=100-1=99

置信区间估计为95%，因为假设总体分布为正态分布，所以拒绝区域对应alpha/2=0.025

查表得出t=1.9842

得出95%的置信区间为-0.00053到0.00007之间

We are 95% confident that the length error is between -0.00053 and 0.00007 inch.

  

b. What assumption must you make about the population distribution in order to construct the confidence interval estimate in (a)?

答：为了构造(a)中的置信区间估计，需要假设总体分布为正态分布。

  

c. Do you think that the assumption needed in order to construct the confidence interval estimate in (a) is valid? Explain.

答：我认为在(a)中构造置信区间估计所需的假设是有效的。

因为样本均值-0.00023英寸，落在95%的置信区间-0.00053到0.00007英寸之间。

所以假设有效。

d. Compare the conclusions reached in (a) with those of Problem 2.41 on page 53

答：教材53页第2.41题，第一问绘制直方图，如下：

汽车公司规定的误差范围是±0.005英寸以内，从直方图的结果来看误差从-0.003英寸开始，一直到0.005英寸（10*0.0008-0.003=0.005），该范围在汽车公司规定的±0.005英寸以内，所以钢厂的工作达到了汽车公司的要求。

根据（a）中得出的结论，与教材题目相比，假设总体分布为正态分布都是成立的。

/*导入数据*/

import excel "C:\Users\86138\Desktop\20231213-发布作业-20231223 1540前提交\Steel.xlsx", sheet("DATA") firstrow clear

/*计算样本均值*/

summarize

/*教材53页第2.41题第一问绘制直方图*/

histogram Error