忒弥斯的蒙眼布 · 2021年12月26日
NO.PZ2015120604000061
问题如下：
According to the above table, what is the standard deviation of Y?
选项：
A.6.64.
B.1.90.
C.2.58.
解释：
C is correct
Expected return of Y = (0.2)(5) + (0.6)(2) + (0.2)(-3) = 1.6
Var(Y) = 0.2(5-1.6)2 + 0.6(2- 1.6)2 + 0.2(-3-1.6)2 = 6.64
Standard deviation= square root of 6.64=2.58
思路都懂，计算器算出来咋不是？用了1、1.2、-0.6三个data，用了2nd 8的键，咋不是标准差这个数值呢
NO.PZ2015120604000061 问题如下 Accorng to the above table, whis the stanrviation of Y? A.6.64. B.1.90. C.2.58. C is correctExpectereturn of Y = (0.2)(5) + (0.6)(2) + (0.2)(-3) = 1.6Var(Y) = 0.2(5-1.6)2 + 0.6(2- 1.6)2 + 0.2(-3-1.6)2 = 6.64Stanrviation= square root of 6.64=2.58 P（x）代表什么
NO.PZ2015120604000061 问题如下 Accorng to the above table, whis the stanrviation of Y? A.6.64. B.1.90. C.2.58. C is correctExpectereturn of Y = (0.2)(5) + (0.6)(2) + (0.2)(-3) = 1.6Var(Y) = 0.2(5-1.6)2 + 0.6(2- 1.6)2 + 0.2(-3-1.6)2 = 6.64Stanrviation= square root of 6.64=2.58 Var(Y) = 0.2(5-1.6)2 + 0.6(2- 1.6)2 + 0.2(-3-1.6)2 = 6.64能用xy或者E（y）这种表示一下吗
NO.PZ2015120604000061 问题如下 Accorng to the above table, whis the stanrviation of Y? A.6.64. B.1.90. C.2.58. C is correctExpectereturn of Y = (0.2)(5) + (0.6)(2) + (0.2)(-3) = 1.6Var(Y) = 0.2(5-1.6)2 + 0.6(2- 1.6)2 + 0.2(-3-1.6)2 = 6.64Stanrviation= square root of 6.64=2.58 里面算法没有用到x的数值，但是这个不仅仅是要的概率。他是联合概率啊，怎么第一步就可以直接用y*联合概率的和就可以得到要expectevalue呢Expectereturn of Y = (0.2)(5) + (0.6)(2) + (0.2)(-3) = 1.6 ？？？？这里面的是联合概率啊，不是单独Y的概率啊
请问表头上joint probabilities 是指Y=5和X=-2同时发生、Y=2 X=1同时发生、Y=-3 X=4同时发生的概率吗？
