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Celestine · 2020年02月01日

问一道题：NO.PZ2015120204000044

  1. CFA II
  2. Quantitative

问题如下：

After applying data transformations, Steele scales the financial data using normalization. She notes that over the full sample dataset, the “Interest Expense” variable ranges from a minimum of 0.2 and a maximum of 12.2, with a mean of 1.1 and a standard deviation of 0.4.

Exhibit 1 Sample of Raw Structured Data Before Cleansing


Based on Exhibit 1, for the firm with ID #3, Steele should compute the scaled value for the “Interest Expense” variable as:

选项：

A.

0.008.

B.

0.083.

C.

0.250.

解释：

B is correct. Steele uses normalization to scale the financial data. Normalization is the process of rescaling numeric variables in the range of [0, 1]. To normalize variable X, the minimum value (Xmin) is subtracted from each observation
(Xi), and then this value is divided by the difference between the maximum and minimum values of X (Xmax – Xmin):
The firm with ID #3 has an interest expense of 1.2. So, its normalized value is calculated as:

Xi(normalized) =(1.2-0.2)/(12.2-0.2)=0.083

这题答案不是（1.2-0.4）/（1.5-0.4）=0.73吗？

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星星_品职助教 · 2020年02月04日

同学你好，

1.5和0.4只是节选出表格里的极值，而真正样本的极值是12.2和0.2

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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？

2021-08-23 07:51 1 · 回答

NO.PZ2015120204000044 这个知识点没有找到，请问一下在讲义的哪个位置

2021-03-06 10:11 1 · 回答

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？

2020-11-07 21:28 1 · 回答

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？

2020-03-01 14:06 2 · 回答

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？

2020-02-18 19:20 1 · 回答