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朱诗怡 · 2020年06月01日

问一道题:NO.PZ2017092702000051

问题如下:

Annual returns and summary statistics for three funds are listed in the following table:

The fund that shows the highest dispersion is:

选项:

A.

Fund PQR if the measure of dispersion is the range.

B.

Fund XYZ if the measure of dispersion is the variance.

C.

Fund ABC if the measure of dispersion is the mean absolute deviation.

解释:

C is correct.

The mean absolute deviation (MAD) of Fund ABC’s returns is greater than the MAD of both of the other funds.

MAD=inXiXnMAD=\frac{\displaystyle\sum_i^n{\vert Xi-\overline X\vert}}n

where \(\overline X\) is the arithmetic mean of the series.

MAD for Fund ABC =

[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\%

MAD for Fund XYZ =

[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}\%

MAD for Fund PQR =

[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}\%

A and B are incorrect because the range and variance of the three funds are as follows:

The numbers shown for variance are understood to be in "percent squared" terms so that when taking the square root, the result is standard deviation in percentage terms. Alternatively, by expressing standard deviation and variance in decimal form, one can avoid the issue of units; in decimal form, the variances for Fund ABC, Fund XYZ, and Fund PQR are 0.0317, 0.0243, and 0.0110, respectively.

答案怎么那么长,难道不是ABC的方差最大,所以选c吗?

1 个答案

星星_品职助教 · 2020年06月02日

同学你好,

方差和range是离散程度的两个不同的衡量指标,仅比较方差没法得到range的结论,所以range也是要比较的。

如果自己有快速方法也不一定就要参照答案。

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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

2021-05-31 00:15 1 · 回答

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

2021-02-16 13:55 1 · 回答

NO.PZ2017092702000051 请问range是什么公式?

2021-02-10 11:50 1 · 回答

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2021-02-06 09:38 1 · 回答

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2020-10-18 12:11 1 · 回答