问题如下图：
不能用standard deviation直接判断吗？
选项：
A. 
B. 
C. 
解释：
FunXYZ if the measure of spersion is the variance. FunAif the measure of spersion is the meabsolute viation. C is correct. The meabsolute viation (MA of FunABC’s returns is greater ththe Mof both of the other fun. MA∑in∣Xi−X‾∣nMA\frac{\splaystyle\sum_i^n{\vert Xi-\overline X\vert}}nMAni∑n∣Xi−X∣ where \(\overline X\) is the arithmetic meof the series. Mfor FunA= [−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%5[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]=14.4% Mfor FunXYZ = [−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%5[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]=9.8% Mfor FunPQR = [−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%5[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]=8.8% A anB are incorrebecause the range anvarianof the three fun are follows: The numbers shown for varianare unrstooto in \"percent square" terms so thwhen taking the square root, the result is stanrviation in percentage terms. Alternatively, expressing stanrviation anvarianin cimform, one cavoithe issue of units; in cimform, the variances for FunABFunXYZ, anFunPQR are 0.0317, 0.0243, an0.0110, respectively. 老师，我用计算器78键上面的数据功能算出的标准差B也对哦，这怎么回事。这个不能用计算器算吗
NO.PZ2017092702000051 FunXYZ if the measure of spersion is the variance. FunAif the measure of spersion is the meabsolute viation. C is correct. The meabsolute viation (MA of FunABC’s returns is greater ththe Mof both of the other fun. MA∑in∣Xi−X‾∣nMA\frac{\splaystyle\sum_i^n{\vert Xi-\overline X\vert}}nMAni∑n∣Xi−X∣ where \(\overline X\) is the arithmetic meof the series. Mfor FunA= [−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%5[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]=14.4% Mfor FunXYZ = [−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%5[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]=9.8% Mfor FunPQR = [−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%5[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]=8.8% A anB are incorrebecause the range anvarianof the three fun are follows: The numbers shown for varianare unrstooto in \"percent square" terms so thwhen taking the square root, the result is stanrviation in percentage terms. Alternatively, expressing stanrviation anvarianin cimform, one cavoithe issue of units; in cimform, the variances for FunABFunXYZ, anFunPQR are 0.0317, 0.0243, an0.0110, respectively.为什么用MA比较离散程度呢
NO.PZ2017092702000051 请问range是什么公式？
FunXYZ if the measure of spersion is the variance. FunAif the measure of spersion is the meabsolute viation. C is correct. The meabsolute viation (MA of FunABC’s returns is greater ththe Mof both of the other fun. MA∑in∣Xi−X‾∣nMA\frac{\splaystyle\sum_i^n{\vert Xi-\overline X\vert}}nMAni∑n∣Xi−X∣ where \(\overline X\) is the arithmetic meof the series. Mfor FunA= [−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]5=14.4%\frac{{\lbrack-20-{(-4)}\rbrack}+{\lbrack23-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}+{\lbrack5-{(-4)}\rbrack}+{\lbrack-14-{(-4)}\rbrack}}5=14.4\%5[−20−(−4)]+[23−(−4)]+[−14−(−4)]+[5−(−4)]+[−14−(−4)]=14.4% Mfor FunXYZ = [−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]5=9.8%\frac{{\lbrack-33-{(-10.8)}\rbrack}+{\lbrack\text{-12}-{(-10.8)}\rbrack}+{\lbrack-\text{12}-{(-10.8)}\rbrack}+{\lbrack\text{-8}-{(-10.8)}\rbrack}+{\lbrack\text{11}-{(-10.8)}\rbrack}}5=\text{9}\text{.8}\%5[−33−(−10.8)]+[-12−(−10.8)]+[−12−(−10.8)]+[-8−(−10.8)]+[11−(−10.8)]=9.8% Mfor FunPQR = [−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]5=8.8%\frac{{\lbrack-\text{14}-{(-\text{5})}\rbrack}+{\lbrack\text{-18}-{(-\text{5})}\rbrack}+{\lbrack\text{6}-{(-\text{5})}\rbrack}+{\lbrack\text{-2}-{(-\text{5})}\rbrack}+{\lbrack\text{3}-{(-\text{5})}\rbrack}}5=\text{8}\text{.8}\%5[−14−(−5)]+[-18−(−5)]+[6−(−5)]+[-2−(−5)]+[3−(−5)]=8.8% A anB are incorrebecause the range anvarianof the three fun are follows: The numbers shown for varianare unrstooto in \"percent square" terms so thwhen taking the square root, the result is stanrviation in percentage terms. Alternatively, expressing stanrviation anvarianin cimform, one cavoithe issue of units; in cimform, the variances for FunABFunXYZ, anFunPQR are 0.0317, 0.0243, an0.0110, respectively.请教一下是否有这样的规律，可以总结出规律，一般A、B两组数，MA的，var也大？对待这类题目这样可以少算一组吗？
根据基础班课程，对比两组或以上数据的离散程度，已知均值和标准差，则用CV就可以判断离散程度的大小？ Fun117.8/4=4.45 Fun215.6/10.8=1.44 Fun310.5/5.0=2.1 所以Fun1离散程度最大 所以在这里也可以基于CV判断，不是非要用十进制方差？ 另外，老师可以把答案中的0.0317、0.0243、0.0110计算过程展示一下吗？
