Calculating μ for a Normal Distribution with Given Probability

[SolCon]

Hello to all.

Simple question but I'm getting the wrong answer here.

Question:

The random variable X is normally distributed with mean μ and variance 21.0. Given that
P(X > 10.0) = 0.7389, find the value of μ.

The equation we'll use is thus: z=(x-μ)/SD
So, what I did was:

0.64 ($$\phi$$0.7389) = (10-μ)/under.root.21
(0.64)(under.root.21)-10=-μ
2.9-10=-μ
-7.1=-μ
μ=7.1

This answer is wrong for some reason and I have no idea why this is so. Any help here is appreciated.

 

[HallsofIvy]

".7389" is the probability that z is less than or equal to .64. You are asked about the probability that a value is larger than 10,

 

[SolCon]

Thanks for the reply.

The answer to this question is 12.9. But this was achieved by having in the above equation "μ-10". Multiplying under root 21 with 0.64 then adding 10 will give us this answer. But isn't the formula "x-μ" instead of "μ-x"?

You have said that the prob being asked is greater than 10. But how will this affect the equation (are you driving at the P(A) = 1- rule? If so, I've done that but that does not bring the correct answer either.) ?