概率论区间估计课件.ppt

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1、推导过程如下:解:有一大批糖果,现从中随机地取16袋,称得重量(克)如下:设袋装糖果的重量服从正态分布,试求总体均值例2附表3-1=0.250.100.050.0250.010.005123456789101112131415161.00000.81650.76490.74070.72670.71760.71110.70640.70270.69980.69740.69550.69380.69240.69120.69013.07771.88561.63771.53321.47591.43981.41491.39681.38301.37221.36341.35621.350

2、21.34501.34061.33686.31382.92002.35342.13182.01501.94321.89461.85951.83311.81251.79591.78231.77091.76131.75311.745912.70624.30273.18242.77642.57062.44692.36462.30602.26222.22812.20102.17882.16042.14482.13152.119931.82076.96464.54073.74693.36493.14272.99802.89652.82142.76382.71812.68102.6

3、5032.62452.60252.583563.65749.92485.84094.60414.03223.70743.49953.35543.24983.16933.10583.05453.01232.97682.94672.9208分布表2.1315就是说估计袋装糖果重量的均值在500.4克与507.1克之间,这个估计的可信程度为95%.这个误差的可信度为95%.解:例3(续例1)如果只假设糖包的重量服从正态分布=0.250.100.050.0250.010.005123456789101112131415161.00000.81650.76490.74070.72

4、670.71760.71110.70640.70270.69980.69740.69550.69380.69240.69120.69013.07771.88561.63771.53321.47591.43981.41491.39681.38301.37221.36341.35621.35021.34501.34061.33686.31382.92002.35342.13182.01501.94321.89461.85951.83311.81251.79591.78231.77091.76131.75311.745912.70624.30273.18242.77642.5

5、7062.44692.36462.30602.26222.22812.20102.17882.16042.14482.13152.119931.82076.96464.54073.74693.36493.14272.99802.89652.82142.76382.71812.68102.65032.62452.60252.583563.65749.92485.84094.60414.03223.70743.49953.35543.24983.16933.10583.05453.01232.97682.94672.92082.2010附表3-2分布表解:例4推导过程如下:

6、根据第五章第二节定理5.4知II.进一步可得:注意:在密度函数不对称时,习惯上仍取对称的分位点来确定置信区间(如图).(续例2)求例2中总体标准差的置信度为0.95的置信区间.解代入公式得标准差的置信区间例5附表4-1=0.250.100.050.0250.010.005123456789101112131415161.3232.7734.1085.3856.6267.8419.03710.21911.38912.54913.70114.84515.98417.11718.24519.3692.7064.6056.2517.7799.23610.64512.01713.

7、36214.68415.98717.27518.54919.81220.06422.30723.5423.8415.9917.8159.48811.07112.59214.06715.50716.91918.30719.67521.02622.36223.68524.99626.2965.0247.3789.34811.14312.83314.44916.01317.53519.02320.48321.92023.33724.73626.11927.48828.8456.6359.21011.34513.27715.08616.81218