问题如下图：
选项：
A. 
B. 
C. 
解释：
老师，像b选项描述的这样，如果99%（2.58）是三倍，95%（1.96）是两倍，那么90%（1.65）应该描述为几倍？ 就是觉得这样描述不准确啊
NO.PZ2017092702000100问题如下 analyst velops the following capitmarket projections.Assuming the returns of the asset classes are scribenormstributions, whiof the following statements is correct?A.Bon have a higher probability of a negative return thstocks.B.On average, 99% of storeturns will fall within two stanrviations of the mean.C.The probability of a bonreturn less thor equto 3% is termineusing a Z-score of 0.25.A is correct.The chanof a negative return falls in the area to the left of 0% unr a stanrnormcurve. stanrzing the returns anstanrviations of the two assets, the likelihooof either asset experiencing a negative return mtermine Z-score (stanrzevalue) = (X – μ)/σ Z-score for a bonreturn of 0% = (0 – 2)/5 = –0.40. Z-score for a storeturn of 0% = (0 – 10)/15 = –0.67. For bon, a 0% return falls 0.40 stanrviations below the mereturn of 2%. In contrast, for stocks, a 0% return falls 0.67 stanrviations below the mereturn of 10%. A stanrviation of 0.40 is less tha stanrviation of 0.67. Negative returns thus occupy more of the left tail of the bonstribution ththe stostribution. Thus, bon are more likely thstocks to experiena negative return.A可以用 标准差/平均 来比较吗C要怎么算呢
NO.PZ2017092702000100问题如下analyst velops the following capitmarket projections.Assuming the returns of the asset classes are scribenormstributions, whiof the following statements is correct?A.Bon have a higher probability of a negative return thstocks.B.On average, 99% of storeturns will fall within two stanrviations of the mean.C.The probability of a bonreturn less thor equto 3% is termineusing a Z-score of 0.25.A is correct.The chanof a negative return falls in the area to the left of 0% unr a stanrnormcurve. stanrzing the returns anstanrviations of the two assets, the likelihooof either asset experiencing a negative return mtermine Z-score (stanrzevalue) = (X – μ)/σ Z-score for a bonreturn of 0% = (0 – 2)/5 = –0.40. Z-score for a storeturn of 0% = (0 – 10)/15 = –0.67. For bon, a 0% return falls 0.40 stanrviations below the mereturn of 2%. In contrast, for stocks, a 0% return falls 0.67 stanrviations below the mereturn of 10%. A stanrviation of 0.40 is less tha stanrviation of 0.67. Negative returns thus occupy more of the left tail of the bonstribution ththe stostribution. Thus, bon are more likely thstocks to experiena negative return.老师请问A要怎么计算
NO.PZ2017092702000100问题如下analyst velops the following capitmarket projections.Assuming the returns of the asset classes are scribenormstributions, whiof the following statements is correct?A.Bon have a higher probability of a negative return thstocks.B.On average, 99% of storeturns will fall within two stanrviations of the mean.C.The probability of a bonreturn less thor equto 3% is termineusing a Z-score of 0.25.A is correct.The chanof a negative return falls in the area to the left of 0% unr a stanrnormcurve. stanrzing the returns anstanrviations of the two assets, the likelihooof either asset experiencing a negative return mtermine Z-score (stanrzevalue) = (X – μ)/σ Z-score for a bonreturn of 0% = (0 – 2)/5 = –0.40. Z-score for a storeturn of 0% = (0 – 10)/15 = –0.67. For bon, a 0% return falls 0.40 stanrviations below the mereturn of 2%. In contrast, for stocks, a 0% return falls 0.67 stanrviations below the mereturn of 10%. A stanrviation of 0.40 is less tha stanrviation of 0.67. Negative returns thus occupy more of the left tail of the bonstribution ththe stostribution. Thus, bon are more likely thstocks to experiena negative return.可以把B答案翻译一下吗？
NO.PZ2017092702000100问题如下 analyst velops the following capitmarket projections.Assuming the returns of the asset classes are scribenormstributions, whiof the following statements is correct?A.Bon have a higher probability of a negative return thstocks.B.On average, 99% of storeturns will fall within two stanrviations of the mean.C.The probability of a bonreturn less thor equto 3% is termineusing a Z-score of 0.25.A is correct.The chanof a negative return falls in the area to the left of 0% unr a stanrnormcurve. stanrzing the returns anstanrviations of the two assets, the likelihooof either asset experiencing a negative return mtermine Z-score (stanrzevalue) = (X – μ)/σ Z-score for a bonreturn of 0% = (0 – 2)/5 = –0.40. Z-score for a storeturn of 0% = (0 – 10)/15 = –0.67. For bon, a 0% return falls 0.40 stanrviations below the mereturn of 2%. In contrast, for stocks, a 0% return falls 0.67 stanrviations below the mereturn of 10%. A stanrviation of 0.40 is less tha stanrviation of 0.67. Negative returns thus occupy more of the left tail of the bonstribution ththe stostribution. Thus, bon are more likely thstocks to experiena negative return.同上，英文版的不太看得懂
NO.PZ2017092702000100 老师这题是不是因为已知了正态分布，所以不用标准化，得出来的概率就可以直接比较大小？
