这里的correlation

NO.PZ2016082405000035 Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243. A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: {l}-2.33=（\frac{-2.33-（0.5）\overline m}{\sqrt{1-0.5^2}}）\\\overline m=-0.62430\enarray}\ 0.31215=−(0.5)m 这一步是怎么来的

NO.PZ2016082405000035 Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243.  A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: l−2.33=−2.33−(0.5)m‾1−0.52−2.33(0.86603)=−2.33−(0.5)m‾0.31215=−(0.5)m‾−0.62430=m‾{l}-2.33=\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}}\\-2.33{(0.86603)}=-2.33-{(0.5)}\overline m\\0.31215=-{(0.5)}\overline m\\-0.62430=\overline ml−2.33=1−0.52 ​−2.33−(0.5)m​−2.33(0.86603)=−2.33−(0.5)m0.31215=−(0.5)m−0.62430=m 请问解答过程中的l-2.33的l是什么意思？

Suppose a cret position ha correlation to the market factor of 0.5. Whis the realizemarket value this useto compute the probability of reaching a fault thresholthe 99% confinlevel? -0.2500. -0.4356. -0.5825. -0.6243.  A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: l−2.33=−2.33−(0.5)m‾1−0.52−2.33(0.86603)=−2.33−(0.5)m‾0.31215=−(0.5)m‾−0.62430=m‾{l}-2.33=\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}}\\-2.33{(0.86603)}=-2.33-{(0.5)}\overline m\\0.31215=-{(0.5)}\overline m\\-0.62430=\overline ml−2.33=1−0.52 ​−2.33−(0.5)m​−2.33(0.86603)=−2.33−(0.5)m0.31215=−(0.5)m−0.62430=m 这讲的是哪一个知识点？ 在那一页ppt啊？

-0.4356. -0.5825. -0.6243.  A fault loss level of 0.01 correspon to -2.33 on the stanrnormstribution. The realizemarket value is computefollows: l−2.33=−2.33−(0.5)m‾1−0.52−2.33(0.86603)=−2.33−(0.5)m‾0.31215=−(0.5)m‾−0.62430=m‾{l}-2.33=\frac{-2.33-(0.5)\overline m}{\sqrt{1-0.5^2}}\\-2.33{(0.86603)}=-2.33-{(0.5)}\overline m\\0.31215=-{(0.5)}\overline m\\-0.62430=\overline ml−2.33=1−0.52 ​−2.33−(0.5)m​−2.33(0.86603)=−2.33−(0.5)m0.31215=−(0.5)m−0.62430=m k是α的临界分位点，α是服从一般正态分布的，为什么k是-2.33？