Suppose that X is a normal Variate with mean 5.

[TomJerry]

Question:
Suppose that X is a normal Variate with mean 5. If P(X>9)=0.2 approximately, what is Var X?

Solution:

Here $$\mu$$=5 and P(X>9) = 0.2

therefore

0.2 = 9 - 5 / V(x)

V(x) = 4 / 0.2 = 20 [Is this correct]

 

[statdad]

Question:
Suppose that X is a normal Variate with mean 5. If P(X>9)=0.2 approximately, what is Var X?

Solution:

Here $$\mu$$=5 and P(X>9) = 0.2

therefore

0.2 = 9 - 5 / V(x)

V(x) = 4 / 0.2 = 20 [Is this correct]

No. 0.2 is a probability. I think I know what you wanted to write when you put

$$\frac{9-5}{V(x)}$$

but, as written, it is a meaningless statement. If you were trying to compare .2 to a Z-score, that won't work: they aren't the same thing.