NO.PZ2015120204000044 0.083. 0.250. B is correct. Steele uses normalization to scale the financitNormalization is the process of rescaling numeric variables in the range of [0, 1]. To normalize variable X, the minimum value (Xmin) is subtractefrom eaobservation (Xi), anthen this value is vithe fferenbetween the maximum anminimum values of X (Xm– Xmin): The firm with I#3 hinterest expense of 1.2. So, its normalizevalue is calculateas: Xi(normalize =(1.2-0.2)/(12.2-0.2)=0.083 解答说要用normalization来做rescaling，那么stanrzation应该什么时候用呢？
NO.PZ2015120204000044 这个知识点没有找到，请问一下在讲义的哪个位置
0.083. 0.250. B is correct. Steele uses normalization to scale the financitNormalization is the process of rescaling numeric variables in the range of [0, 1]. To normalize variable X, the minimum value (Xmin) is subtractefrom eaobservation (Xi), anthen this value is vithe fferenbetween the maximum anminimum values of X (Xm– Xmin): The firm with I#3 hinterest expense of 1.2. So, its normalizevalue is calculateas: Xi(normalize =(1.2-0.2)/(12.2-0.2)=0.083 请问这道题涉及的知识点是？
0.083. 0.250. B is correct. Steele uses normalization to scale the financitNormalization is the process of rescaling numeric variables in the range of [0, 1]. To normalize variable X, the minimum value (Xmin) is subtractefrom eaobservation (Xi), anthen this value is vithe fferenbetween the maximum anminimum values of X (Xm– Xmin): The firm with I#3 hinterest expense of 1.2. So, its normalizevalue is calculateas: Xi(normalize =(1.2-0.2)/(12.2-0.2)=0.083 为什么不是（1.2-1.1）/0.4=0.25？
0.083. 0.250. B is correct. Steele uses normalization to scale the financitNormalization is the process of rescaling numeric variables in the range of [0, 1]. To normalize variable X, the minimum value (Xmin) is subtractefrom eaobservation (Xi), anthen this value is vithe fferenbetween the maximum anminimum values of X (Xm– Xmin): The firm with I#3 hinterest expense of 1.2. So, its normalizevalue is calculateas: Xi(normalize =(1.2-0.2)/(12.2-0.2)=0.083 是因为题干中没说是正态分布，所以不用mean和标准差求解吗？
