Reverse Z-Value for normal distrubtion Wondering if I'm on the right path on this question. 

Let $X$ be a Normal distributed stochastic value with $\mu=4$ and $\sigma = 7$ Find a value $C$ so that $P(X>C)=0.78$

The way I've solved this question was to first change the equation like so
$$1-P(X<C)=0.78$$
$$P(X<C)=0.22$$
Then find the $Z$ value that equals $0.22$ and that would be $Z(-0.77)$ That makes my new equation equal to.
$$P(X<C)=Z(-0.77)$$
I then use the formula for the normal distrubtion and solve for a C that would make the equation equal to $-0.77$
$$\frac{C-4}{7}=-0.77$$
And then I get the answer $$C=-1.39$$
I would believe my execution is right but it differs from the solution my teacher has given. Granted his solution set has a couple of miscalculations because it has not been verified I have yet to find anyone complain about his answer on this one which leads me to believe that I have made an error. 
His answer was, by the way, $$C=9.39$$
 A: This is indeed a question about the inverse of a normal cumulative distribution function (CDF). Such inverse functions are often called quantile functions. The particulars of the computation depend on whether you are using software or printed normal tables.
Software: If you have software available, you can use your equation $P(X \le C) = 0.22,$
where $X \sim \mathsf{Norm}(\mu = 4,\, \sigma = 7).$ In R statistical
software normal quantile functions are denoted qnorm, so the answer $-1.405$ can
be obtained as:
qnorm(.22, 4, 7)
# [1] -1.405352

This is similar to @Henry's code, in which the argument lower.tail=F
takes 22% of the probability from the upper tail, whereas qnorm
without this argument takes 22% from the lower tail (according to the
usual definition of the CDF). 
Other statistical and mathematical software and calculators perform
similar computations using different syntax.
The plot below shows the PDF and CDF of $\mathsf{Norm}(4, 7)$ with
red dashed lines indicating key quantities.

Printed tables: If you doing this with printed standard normal CDF tables,
then you need to 'standardize' much as you have done:
$$P(X \le C) = P\left(\frac{X - \mu}{\sigma} \le \frac{C - 4}{7}\right)
= P\left(Z \le \frac{C-4}{7}\right) = 0.22,$$
where $Z \sim \mathsf{Norm}(0,1).$ Looking in the body of the table
you can find a z-score that corresponds as nearly as possible to 0.22
(where the meaning of 'corresponds' depends on the format of the table). Here, the corresponding
'z-score' in the margin of the table is $-0.77.$ (Most tables allow no
more than two-place accuracy). 
Then $(C - 4)/7 = -0.77)$ so that $C \approx -1.39.$ But you can trust only two significant digits of the answer,
so it amounts to $-1.4.$ By interpolation in the body of the table, you
might get a little more accuracy. But your answer is about as close
as you can reliably get using printed normal tables.
