Normal Distribution finding values The question says:

X is normal with mean -1 and variance 4. Find the value $x_0$ for which the probability is $.2676$ that $X$ will take on a value less than $x_0$.

I know this has to deal with normal probability distribution. I am trying to learn these types of problems, but I am stuck on this one. Can someone help me?
I figured that $x_0$ is the $26.76^{th}$ percentile. I also know we might have to use the z-chart tables. 
The formulas that came to mind are $Z=X-\mu/\sigma$. We can turn that into $\mu=X-Z\sigma$.
I did $P(X \leq x_0)=.2676$
On a normal distribution curve this would be $.5000-.2676=.2324$
This would give us a $Z$-score of $0.62$
I then use $Z=(X-\mu)/\sigma$ $=(X-(-1))/4=1.48$ which would give us an answer, but it does not seem correct. 
Can some help me with the answer and how to show the working?
 A: One can also use the negative Z-scores. Please look at the  negative Z score table from http://clas.sa.ucsb.edu/staff/binh/stdNormalTable.pdf. 
From the negative Z-score table we can get $z_0$ of $Pr[Z \leq z_0]=0.2676$, which is $z_0=-0.62$. Then
$Pr[Z \leq z_{0}]=Pr[X \leq x_0]$, where $~x_0=(z_0 *\sigma)+\mu$ 
A: First note that $X$ has a distribution symmetric about $-1$, so our point $x_0$ will be to the left of $-1$. 
We have 
$$\Pr(X\le x_0)=\Pr\left(Z\le \frac{x_0-(-1)}{\sqrt{4}}\right)=0.2676.$$
We want the place $-w$ such that the probability that $Z\le -w$ is $0.2676$.
This information is not directly available from the usual tables, since $-w$ is negative. 
But we want $\Pr(Z\ge -w)=0.7324$. By symmetry, we want $\Pr(Z\le w)=0.7324$. Tables tell us $w\approx 0.62$. Thus
$$\frac{x_0+1}{2}\approx -0.62.$$
Now we can find $x_0$.   
Remark: Your $z$-score was not quite right, it should have been $-0.62$. And then there was a slip, you used the variance $4$ instead of the standard deviation $2$. The most important thing is to realize from the beginning what is in the first sentence of the answer. We are dealing with the left tail of the distribution. 
