standard deviation 是不是算错了？var(x)开完根号不应该等于0.02吗？

NO.PZ2020010302000010 问题如下 Suppose the return on asset hthe following stribution:Compute the mean, variance, anstanrviation.Verify your result in (computing E[X2]E[X^2]E[X2] rectly anusing the alternative expression for the variance.Is this stribution skewe es this stribution have excess kurtosis? e. Whis the meof this stribution? The meis E[X] = Σx Pr(X = x) = 0.25%.The varianis Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555Var[X] = Σ(x - E[X])^2 Pr(X = x) = 0.000555Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555.The stanrviation is Var[X]=2.355\sqrt {Var[X]} = 2.355%Var[X]​=2.355.E[X2]=Σx2Pr(X=x)=.000561E[X^2] = Σx^2 Pr(X = x) = .000561E[X2]=Σx2Pr(X=x)=.000561 anso E[X2]−(E[X])2=0.000561−(.0025)2=.000555E[X^2] - (E[X])^2 = 0.000561 - (.0025)^2 = .000555E[X2]−(E[X])2=0.000561−(.0025)2=.000555, whiis the same.The skewness requires computingskew(X)=E[X−E[X]]3/σ3=E[(X−μσ)3]=Σ(x−μσ)3Pr(X−x)skew(X)=E[X-E[X]]^3/{\sigma^3}=E[(\frac{X-\mu}{\sigma})^3]=Σ(\frac{x-\mu}{\sigma})^3Pr(X-x)skew(X)=E[X−E[X]]3/σ3=E[(σX−μ​)3]=Σ(σx−μ​)3Pr(X−x)Thus the skewness is 0.021, anthe stribution ha milpositive skew. The kurtosis requires computingkurtosis(X)=E[(X−E[X])4]σ4=E[(X−μσ)4]=Σ(x−μσ)4Pr(X−x)kurtosis(X)=\frac{E[(X-E[X])^4]}{\sigma^4}=E[(\frac{X-\mu}{\sigma})^4]=Σ(\frac{x-\mu}{\sigma})^4Pr(X-x)kurtosis(X)=σ4E[(X−E[X])4]​=E[(σX−μ​)4]=Σ(σx−μ​)4Pr(X−x)Thus the kurtosis is 2.24. The excess kurtosis is then 2.24 - 3 = -0.76. This stribution es not have excess kurtosis.e. The meis the value where least 50% probability lies to the left, anleast 50% probability lies to the right. Cumulating the probabilities into a C, this occurs the return value of 0%. 请问带概率也可以计算机计算吗？如何计算？计算器能否计算kurtosis、skewness

NO.PZ2020010302000010问题如下 Suppose the return on asset hthe following stribution:Compute the mean, variance, anstanrviation.Verify your result in (computing E[X2]E[X^2]E[X2] rectly anusing the alternative expression for the variance.Is this stribution skewe es this stribution have excess kurtosis? e. Whis the meof this stribution?The meis E[X] = Σx Pr(X = x) = 0.25%.The varianis Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555Var[X] = Σ(x - E[X])^2 Pr(X = x) = 0.000555Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555.The stanrviation is Var[X]=2.355\sqrt {Var[X]} = 2.355%Var[X]​=2.355.E[X2]=Σx2Pr(X=x)=.000561E[X^2] = Σx^2 Pr(X = x) = .000561E[X2]=Σx2Pr(X=x)=.000561 anso E[X2]−(E[X])2=0.000561−(.0025)2=.000555E[X^2] - (E[X])^2 = 0.000561 - (.0025)^2 = .000555E[X2]−(E[X])2=0.000561−(.0025)2=.000555, whiis the same.The skewness requires computingskew(X)=E[X−E[X]]3/σ3=E[(X−μσ)3]=Σ(x−μσ)3Pr(X−x)skew(X)=E[X-E[X]]^3/{\sigma^3}=E[(\frac{X-\mu}{\sigma})^3]=Σ(\frac{x-\mu}{\sigma})^3Pr(X-x)skew(X)=E[X−E[X]]3/σ3=E[(σX−μ​)3]=Σ(σx−μ​)3Pr(X−x)Thus the skewness is 0.021, anthe stribution ha milpositive skew. The kurtosis requires computingkurtosis(X)=E[(X−E[X])4]σ4=E[(X−μσ)4]=Σ(x−μσ)4Pr(X−x)kurtosis(X)=\frac{E[(X-E[X])^4]}{\sigma^4}=E[(\frac{X-\mu}{\sigma})^4]=Σ(\frac{x-\mu}{\sigma})^4Pr(X-x)kurtosis(X)=σ4E[(X−E[X])4]​=E[(σX−μ​)4]=Σ(σx−μ​)4Pr(X−x)Thus the kurtosis is 2.24. The excess kurtosis is then 2.24 - 3 = -0.76. This stribution es not have excess kurtosis.e. The meis the value where least 50% probability lies to the left, anleast 50% probability lies to the right. Cumulating the probabilities into a C, this occurs the return value of 0%.根据老师的讲解， 中位数应该是第5个数和第6个数中间啊，不是0.5%吗，为什么是0？

NO.PZ2020010302000010 问题如下 Suppose the return on asset hthe following stribution:Compute the mean, variance, anstanrviation.Verify your result in (computing E[X2]E[X^2]E[X2] rectly anusing the alternative expression for the variance.Is this stribution skewe es this stribution have excess kurtosis? e. Whis the meof this stribution? The meis E[X] = Σx Pr(X = x) = 0.25%.The varianis Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555Var[X] = Σ(x - E[X])^2 Pr(X = x) = 0.000555Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555.The stanrviation is Var[X]=2.355\sqrt {Var[X]} = 2.355%Var[X]​=2.355.E[X2]=Σx2Pr(X=x)=.000561E[X^2] = Σx^2 Pr(X = x) = .000561E[X2]=Σx2Pr(X=x)=.000561 anso E[X2]−(E[X])2=0.000561−(.0025)2=.000555E[X^2] - (E[X])^2 = 0.000561 - (.0025)^2 = .000555E[X2]−(E[X])2=0.000561−(.0025)2=.000555, whiis the same.The skewness requires computingskew(X)=E[X−E[X]]3/σ3=E[(X−μσ)3]=Σ(x−μσ)3Pr(X−x)skew(X)=E[X-E[X]]^3/{\sigma^3}=E[(\frac{X-\mu}{\sigma})^3]=Σ(\frac{x-\mu}{\sigma})^3Pr(X-x)skew(X)=E[X−E[X]]3/σ3=E[(σX−μ​)3]=Σ(σx−μ​)3Pr(X−x)Thus the skewness is 0.021, anthe stribution ha milpositive skew. The kurtosis requires computingkurtosis(X)=E[(X−E[X])4]σ4=E[(X−μσ)4]=Σ(x−μσ)4Pr(X−x)kurtosis(X)=\frac{E[(X-E[X])^4]}{\sigma^4}=E[(\frac{X-\mu}{\sigma})^4]=Σ(\frac{x-\mu}{\sigma})^4Pr(X-x)kurtosis(X)=σ4E[(X−E[X])4]​=E[(σX−μ​)4]=Σ(σx−μ​)4Pr(X−x)Thus the kurtosis is 2.24. The excess kurtosis is then 2.24 - 3 = -0.76. This stribution es not have excess kurtosis.e. The meis the value where least 50% probability lies to the left, anleast 50% probability lies to the right. Cumulating the probabilities into a C, this occurs the return value of 0%. 请问一下这类型的题目是应该使用计算器么？如果是的话应该怎么输入呢？谢谢。

NO.PZ2020010302000010 问题如下 Suppose the return on asset hthe following stribution:Compute the mean, variance, anstanrviation.Verify your result in (computing E[X2]E[X^2]E[X2] rectly anusing the alternative expression for the variance.Is this stribution skewe es this stribution have excess kurtosis? e. Whis the meof this stribution? The meis E[X] = Σx Pr(X = x) = 0.25%.The varianis Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555Var[X] = Σ(x - E[X])^2 Pr(X = x) = 0.000555Var[X]=Σ(x−E[X])2Pr(X=x)=0.000555.The stanrviation is Var[X]=2.355\sqrt {Var[X]} = 2.355%Var[X]​=2.355.E[X2]=Σx2Pr(X=x)=.000561E[X^2] = Σx^2 Pr(X = x) = .000561E[X2]=Σx2Pr(X=x)=.000561 anso E[X2]−(E[X])2=0.000561−(.0025)2=.000555E[X^2] - (E[X])^2 = 0.000561 - (.0025)^2 = .000555E[X2]−(E[X])2=0.000561−(.0025)2=.000555, whiis the same.The skewness requires computingskew(X)=E[X−E[X]]3/σ3=E[(X−μσ)3]=Σ(x−μσ)3Pr(X−x)skew(X)=E[X-E[X]]^3/{\sigma^3}=E[(\frac{X-\mu}{\sigma})^3]=Σ(\frac{x-\mu}{\sigma})^3Pr(X-x)skew(X)=E[X−E[X]]3/σ3=E[(σX−μ​)3]=Σ(σx−μ​)3Pr(X−x)Thus the skewness is 0.021, anthe stribution ha milpositive skew. The kurtosis requires computingkurtosis(X)=E[(X−E[X])4]σ4=E[(X−μσ)4]=Σ(x−μσ)4Pr(X−x)kurtosis(X)=\frac{E[(X-E[X])^4]}{\sigma^4}=E[(\frac{X-\mu}{\sigma})^4]=Σ(\frac{x-\mu}{\sigma})^4Pr(X-x)kurtosis(X)=σ4E[(X−E[X])4]​=E[(σX−μ​)4]=Σ(σx−μ​)4Pr(X−x)Thus the kurtosis is 2.24. The excess kurtosis is then 2.24 - 3 = -0.76. This stribution es not have excess kurtosis.e. The meis the value where least 50% probability lies to the left, anleast 50% probability lies to the right. Cumulating the probabilities into a C, this occurs the return value of 0%. The stanrviation 不应该是2.356%吗为什么是2.355