Height : Normal distribution I am looking at the following: 
The average tallest men live in Netherlands and Montenegro mit $1.83$m=$183$cm. 
The average shortest men live in Indonesia mit $1.58$m=$158$cm. 
The standard deviation of the height in Netherlands/Montenegro is $9.7$cm and in Indonesia it is $7.8$cm. 
The height of a giant of Indonesia is exactly 2 standard deviations over the average height of an Indonesian. He goes to Netherlands. Which is the part of the Netherlands that are taller than that giant? 
I thought to do the following: 
Since the height of a giant of Indonesia is exactly 2 standard deviations over the average height of an Indonesian, we get that his height is $158+2\cdot 7.8=173.6$cm, right? 
We have the following: 

Now we want to compute $P(x>173.6)=1-P(x\leq 173.6)$, right?
At the graph we have $173.3$ how could we compute the $P(x\leq 173.6)$ ? 
 A: You are right. $X$ is distributed as $\mathcal N(183, 9.7^2)$. To compute $P(X\leq 173.6)$ you use the standardized radom variable $Z=\frac{X-\mu}{\sigma}$, where $Z\sim \mathcal N(0,1)$
$P(X\leq 173.6)=\Phi\left(\frac{173.6-183}{9.7}\right)\approx\Phi(-0.97)$
$\Phi(z)$ is the cdf of the standard normal distribution. You can look at this table what $\Phi(-0.97)$ is. For orientation, the value is between $14\%$ and $18\%$.
Link to a online calculator.

Let mm be the minimal acceptable height, then $P(x>m)=0,01$, or not?
  It also equivalent to $P(x≤m)=0.99$, right?

You are right that both equations are equivalent. You have made the right transformations.

$\frac{m-158}{7.8}=2.32 \Rightarrow m=176.174\ cm$ Is this correct?

More or less. We usually say that $\Phi(2.33)=0.99$. It is $\Phi(2.32)=0.98983$ and $\Phi(2.33)=0.99010$. The second value is nearer to 0.9 than the first value. 
But the funny thing is that if I use $2.33$ the result is $m=176.174$. Maybe you have used 2.33 on the RHS.
Your answer to the second question is right. $\large \checkmark$
