VAR std deviation

We need a result from the tables of 0.4. 1.28 gives 0.3997 and 1.29 gives 0.4015, and therefore to get 0.4 is between the two and we just apportion between them assuming there is a linear relationship. The difference between 0.3997 and 0.4015 is 0.0018. To get 0.4 we need to be 0.0003 above 0.3997, and so apportioning gives an answer of 1.28 + (0.0003/0.0018 x (1.29 - 1.28)) = 1.282 However, I would not spend time apportioning and would just choose the nearest (in the case 1.28). I do not believe that taking 1.28 would lose any marks at all :-)