For a normal random variable X with parameters $\mu =10$ and $\sigma^2 = 36$
Find $$P(X>5)= P(\frac{X-10}{6} \lt \frac{5-10}{6})= P(Z > \frac{-5}{6})= 1-\phi(\frac{-5}{6})=\phi (\frac{5}{6}).$$
$...=?$
on a side note
$Z = \frac{5-10}{6}$
I dont have much experience with these types of problems so if I am way off any help would be appreciated
You have a lot of typos so I'll write out the calculation here and explain it as I go. First,
$$ \begin{align} P(X>5) &= P\left(\frac{X-10}{6} > \frac{5-10}{6}\right) \\&= P\left(Z > -5/6\right)\end{align}$$
where in the first step we just rewrote the inequality and in the second we defined $Z=(X-10)/6.$ (Not $(5-10)/6$.)
Since $X$ is a normal with mean $10$ and standard deviation $6,$ $Z$ is a normal with mean $0$ and standard deviation $1.$
We can then write $$ P(Z>-5/6) = 1-P(Z\le -5/6).$$
This just says that the probability of $Z$ being greater than $-5/6$ is equal to one minus the probability of it not being greater than $-5/6.$ These are the only two possibilities so their probabilities must add up to one.
Now we can use the fact that the cumulative distribution function for the standard normal, usually denoted $\Phi(x)$, is defined as $$\Phi(x) = P(Z \le x).$$
Plugging this in we have $$ P(X>5) = 1-\Phi(-5/6).$$
Lastly, the cumulative for the standard normal (actually the CDF for any continuous random variable that's distributed symmetrically around zero) obeys the identity $$ 1-\Phi(-x) = \Phi(x).$$ This can be seen by drawing a bell curve and sketching the area that each side represents. The left hand side represents everything greater than $-x$ and the right hand side represents everything less than $x.$ From symmetry these are equal.
Plugging in this last identity gives $$ P(X>5) = 1-\Phi(-5/6) = \Phi(5/6).$$
To compute this, you need to compute $$ \int_{-\infty}^{5/6}\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx$$ by some means. Oftentimes in beginning classes you're given tables for $\Phi$ and can look up the value at $5/6$ from there. Also, you can use mathematica/ wolfram alpha or a software package that has the normal cumulative built in. I typed "normcdf(5/6)" into an Octave prompt and got 0.79767. Wolfram alpha gives the same thing. If you're using tables you can check against this number to make sure you're doing it right.
You are on the right track. A small correction, $Z = \frac{X-10}{6} \not = \frac{5-10}{6}$, because $Z$ is a random variable, not a constant.
The following is pretty much what you wrote, but with a few corrections.
$$\mathbb{P}(X > 5) = 1 - \mathbb{P}(X < 5) = 1 - \mathbb{P}(\frac{X-10}{6} < \frac{5-10}{6}) = 1 -\mathbb{P}(Z < \frac{-5}{6})$$
And now you can either look that up in a z-table or ask a computer or calculator to find it for you.