问一道题：NO.PZ2020010303000011

NO.PZ2020010303000011问题如下 If the return on a stock, R, is normally stributewith a ily meof 8%/252 ana ily varianof (20%)2/252(20\%)^2/252(20%)2/252, finthe values where Pr(R r) = .001Pr(R r) = .01Pr(R r) = .05 The meis 0.031% per y anthe varianis 1.58 per y (so ththe stanrviation is 1.26% per y). To finthese values, we transform the variable to stanrnormal, so thatPr(R r)=.001=Pr(Z r−μσ)=.001 Pr(R r) = .001 = Pr(Z \frac{r-\mu}{\sigma})= .001 Pr(R r)=.001=Pr(Z σr−μ​)=.001The value for the stanrnormis -3.09(NORM.S.INV(0.001) in Excel) so th−3.09∗σ+μ=−3.86%-3.09 * \sigma + \mu = -3.86\%−3.09∗σ+μ=−3.86%.The same ia usehere where z = -2.33 so thPr(Z z) = .01. Transforming this value, r=−2.33∗σ+μ=−2.89%r = -2.33 * \sigma + \mu = -2.89\%r=−2.33∗σ+μ=−2.89%.Here the value of z is -1.645 so thr=−1.645∗σ+μ=−2.04%r = -1.645 * \sigma + \mu = -2.04\%r=−1.645∗σ+μ=−2.04%.These are all common Vquan-tiles ansuggest ththere is a 5% chanththe return woulless th-2.04% on any given y, a 1% change thit woulless th-2.89%, ana one in 1,000 chanththe return woulless th-3.86%, if returns were normally stribute 老师好，方差是用20%^2/252吗？这样算，得出的是方差是0.0158%，而不是1.58%。另外，20%^2大于252，能直接用20%^2除以252吗？

NO.PZ2020010303000011 问题如下 If the return on a stock, R, is normally stributewith a ily meof 8%/252 ana ily varianof (20%)2/252(20\%)^2/252(20%)2/252, finthe values where Pr(R r) = .001Pr(R r) = .01Pr(R r) = .05 The meis 0.031% per y anthe varianis 1.58 per y (so ththe stanrviation is 1.26% per y). To finthese values, we transform the variable to stanrnormal, so thatPr(R r)=.001=Pr(Z r−μσ)=.001 Pr(R r) = .001 = Pr(Z \frac{r-\mu}{\sigma})= .001 Pr(R r)=.001=Pr(Z σr−μ​)=.001The value for the stanrnormis -3.09(NORM.S.INV(0.001) in Excel) so th−3.09∗σ+μ=−3.86%-3.09 * \sigma + \mu = -3.86\%−3.09∗σ+μ=−3.86%.The same ia usehere where z = -2.33 so thPr(Z z) = .01. Transforming this value, r=−2.33∗σ+μ=−2.89%r = -2.33 * \sigma + \mu = -2.89\%r=−2.33∗σ+μ=−2.89%.Here the value of z is -1.645 so thr=−1.645∗σ+μ=−2.04%r = -1.645 * \sigma + \mu = -2.04\%r=−1.645∗σ+μ=−2.04%.These are all common Vquan-tiles ansuggest ththere is a 5% chanththe return woulless th-2.04% on any given y, a 1% change thit woulless th-2.89%, ana one in 1,000 chanththe return woulless th-3.86%, if returns were normally stribute 请问这是已知概率，反求z的取值吗？查表方式可以计算？

NO.PZ2020010303000011 问题如下 If the return on a stock, R, is normally stributewith a ily meof 8%/252 ana ily varianof (20%)2/252(20\%)^2/252(20%)2/252, finthe values where Pr(R r) = .001Pr(R r) = .01Pr(R r) = .05 The meis 0.031% per y anthe varianis 1.58 per y (so ththe stanrviation is 1.26% per y). To finthese values, we transform the variable to stanrnormal, so thatPr(R r)=.001=Pr(Z r−μσ)=.001 Pr(R r) = .001 = Pr(Z \frac{r-\mu}{\sigma})= .001 Pr(R r)=.001=Pr(Z σr−μ​)=.001The value for the stanrnormis -3.09(NORM.S.INV(0.001) in Excel) so th−3.09∗σ+μ=−3.86%-3.09 * \sigma + \mu = -3.86\%−3.09∗σ+μ=−3.86%.The same ia usehere where z = -2.32 so thPr(Z z) = .01. Transforming this value, r=−2.32∗σ+μ=−2.89%r = -2.32 * \sigma + \mu = -2.89\%r=−2.32∗σ+μ=−2.89%.Here the value of z is -1.645 so thr=−1.645∗σ+μ=−2.04%r = -1.645 * \sigma + \mu = -2.04\%r=−1.645∗σ+μ=−2.04%.These are all common Vquan-tiles ansuggest ththere is a 5% chanththe return woulless th-2.04% on any given y, a 1% change thit woulless th-2.89%, ana one in 1,000 chanththe return woulless th-3.86%, if returns were normally stribute

老师，这道题你了，但是我没有理解，请详细下这套题的分析过程。

a ily varianof (20%)²>252 怎么推出“the varianis 1.58 per y (so ththe stanrviation is 1.26% per y)”？