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.

$$\mathbb{P}[X>5600]=8\%=0.08$$
$$\mathbb{P}[X\leq 5600]=92\%=0.92$$
that is
$$\Phi\Bigg[\frac{5600-\mu}{392}\Bigg]=0.92$$
Now using Z-table you get
$$\frac{5600-\mu}{392}=1.4051$$
that is
$$\mu=5049$$

How to read a Z-table

I used a calculator. With Paper -table you get that the quantile corresponding to the probability of $0.92$ is between $1.40$ and $1.41$, more or less exactly in the middle...
