问题如下图:
选项:
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解释:
NO.PZ2016062402000012 请解析答案,认真听了课,但是完全不会这个题。
$70.80 an$149.20 $74.68 an$163.56 $102.18 an$119.53 Note ththis is a two-taileconfinban so thα = 1.96. We finthe extreme values from $100 eμ±ασe^{\mu\pm\alpha\sigma}eμ±ασ The lower limit is then V1=$100e0.10−1.96×0.2=$100e−0.292=$74.68V_1=\$100e^{0.10-1.96\times0.2} =\$100e^{-0.292}=\$74.68V1=$100e0.10−1.96×0.2=$100e−0.292=$74.68. The upper limit is V2=$100e0.10+1.96×0.2=$100e0.492=$163.56V_2=\$100e^{0.10+1.96\times0.2}=\$100e^{0.492}=\$163.56V2=$100e0.10+1.96×0.2=$100e0.492=$163.56. 老师能详细一下这道题吗
老师您好,突然有个疑问,那几个特殊点对应的区间是针对于标准正态分布的还是正态分布?谢谢
这里写的是α= 1.96,正确的表达是不是“α =5%时,双尾检验下的z= 1.96”