finding the mean in normal distribution I am struggling on this question where I need to find the mean given the standard deviation.
The question is the following:
A certain brand of flood lamps has a length of life that is normally distributed, 
with a standard deviation of 392 hours.
If 8 % of the lamps last more than 5600 hours, what is the mean length of life? 

This is what I've tried so far
let $X$ = length of light
$X\sim\mathcal{N}(\text{mean = unknown }(u), SD = 392)$
$P(X > 5600) = 0.8$
$1 - P(X < 5600) = 0.8$ (stuck after this part)
$1 - P(5600-u/392 < 5600) = 0.8$ ( incorrect)
I am having troubles working backwards to the find the mean.
Could someone point me in the right direction.

\begin{align}
X &\sim \mathcal{N}(\mu, 392^2) \\
P(X \le 5600) &= 0.92 \\
P \left( Z \le \frac{5600 - \mu}{392} \right) &= 0.92 \tag{normalize} \\
\mu \approx 5600 - 392 (1.41) &= 5047.28 \tag{check standard normal table}
\end{align}
P.S. My answer used $\approx$ instead of $=$ as the value is only an approximation.
