remoo · 2020年01月20日
问题如下：
A stock obeying normal distribution has a mean return of 6% and a standard deviation of 18%. The probability for the stock's return to fall less than 4% is:
选项：
A.37.91%
B.45.62%
C.57.68%
解释：
B is correct. Firstly, standardize the original normal distribution：Z= (x-6%)/18%, so P(x <4%) = P( Z<-0.11).
Notice that N(-0.11) = 1- N(0.11). Based on the given table, N(0.11) = 0.5438, so N(-0.11) = 1-0.5438 = 0.4562 = 45.62%
Z在计算时 不是要除以标准误吗？即sigama/根下N？
NO.PZ2018062016000088问题如下 A stoobeying normstribution ha mereturn of 6% ana stanrviation of 18%. The probability for the stock's return to fall less th4% is:N(0.11) = 0.5438 A.37.91%B.45.62%C.57.68% B is correct. Firstly, stanrze the originnormstributionZ= (x-6%)/18%, so P(x 4%) = P( Z -0.11).NotithN(-0.11) = 1- N(0.11). Baseon the given contion, N(0.11) = 0.5438, so N(-0.11) = 1-0.5438 = 0.4562 = 45.62% 另外题目给的是N也是累积概率的意思吗？为什么给的不是F
NO.PZ2018062016000088问题如下 A stoobeying normstribution ha mereturn of 6% ana stanrviation of 18%. The probability for the stock's return to fall less th4% is:N(0.11) = 0.5438 A.37.91%B.45.62%C.57.68% B is correct. Firstly, stanrze the originnormstributionZ= (x-6%)/18%, so P(x 4%) = P( Z -0.11).NotithN(-0.11) = 1- N(0.11). Baseon the given contion, N(0.11) = 0.5438, so N(-0.11) = 1-0.5438 = 0.4562 = 45.62% 为什么能看出来是FX、有没有可能是T分布、0.11指的就是criticpoint右边区域的面积呢？
NO.PZ2018062016000088 45.62% 57.68% B is correct. Firstly, stanrze the originnormstributionZ= (x-6%)/18%, so P(x <4%) = P( Z<-0.11). NotithN(-0.11) = 1- N(0.11). Baseon the given contion, N(0.11) = 0.5438, so N(-0.11) = 1-0.5438 = 0.4562 = 45.62% 这道题哪句话说的是Z分布，另外x为什么是4呢？不太明白这道题目的意思
45.62% 57.68% B is correct. Firstly, stanrze the originnormstributionZ= (x-6%)/18%, so P(x <4%) = P( Z<-0.11). NotithN(-0.11) = 1- N(0.11). Baseon the given table, N(0.11) = 0.5438, so N(-0.11) = 1-0.5438 = 0.4562 = 45.62%请问考试时如何查表呢？
老师那个0.11是用 （4-6）/18得到的吗？
