How to find $z$-score I have some probabilities, but I have to find the $z$-score. I am not sure how do to this when I am told I have to use slope-intercept. Where do I plug the numbers in exactly?
Here is one of my problems:

Find $d^{\prime}$ and locate $X_C$ approximately on drawing of
distributions.
$$\begin{array}{|l|l|l|} \hline \text{Response} & \text{Stimuli}\\ & N
 & S+N  \\ \hline N & 39 & 21 \\ S+N & 30 & 57 \\ \hline \end{array}$$


First step is to convert raw numbers into probabilities
$$
\begin{array}{lclcl}
p(\text{HIT}) &=& p(Y|S+N) = 57/(57+21) = 57/78 &=& 0.7308 \\
p(\text{FA}) &=& p(Y|N) = 30/(39+30) = 30/69 &=& 0.4347
\end{array}
$$
You then need to use the table to convert these values to $z$-scores. Remember because the table does not have every value, you will need to use a slope intercept approach to calculate this value.


Then use the formula to calculate $d'$.

EDIT: Here is the table it is referring to:
\begin{array}{|}
\hline
\text{$\quad\quad$Tabled values of the normal curve}\\ \hline
 \text{area}\ 0-t & t\\ 
 0  & 0\\ 
 0.39  & 0.1 \\ 
 0.79 &0.2 \\ 
 0.118  & 0.3\\ 
 0.155  & 0.4\\ 
 0.192 & 0.5\\ 
 0.226  & 0.6\\ 
 0.258  & 0.7\\ 
 0.288  & 0.8\\ 
 0.316 & 0.9\\ 
 0.341  & 1.0\\ 
 0.364  & 1.1\\ 
 0.385  & 1.2\\ 
 0.403  & 1.3\\ 
 0.419  & 1.4\\ 
 0.433  & 1.5\\ 
 0.445 & 1.6\\ 
 0.455  & 1.7\\ 
 0.464 & 1.8\\ 
 0.471  & 1.9\\
 0.477  & 2.0\\
 0.482  & 2.1\\ 
 0.486  & 2.2\\ 
 0.489  & 2.3\\ 
 0.492  & 2.4\\ 
 0.494  & 2.5\\ \hline
\end{array}
 A: The table you've linked is a pretty nonstandard format for a z-score table, but it seems to be referring to a situation like this:

The area under the curve between $0$ and $t$ is the probability of a normally distributed variable falling between $0$ and $t$. Using the fact that the curve is symmetric about $0$, you can deduce the probability of a normally distributed variable falling in any interval you care to.
A $z$-score would correspond to $t$ on the diagram and in your table. However, the probability associated with a $z$-score is not the probability of falling into $[0, t]$ (as appears on the table), but rather the cumulative probability, that is, the probability of falling into $[-\infty, t]$. Since you're finding cumulative probabilities in your first step (at least, I sure hope you are because if you're directly finding probabilities for $[0, t]$ then something like $.7308$ makes no sense) you need an extra step, to go from the full cumulative probability to the probability seen on the table.
If your cumulative probability is greater than $.5$ (i.e., $t > 0$), all you need to do is subtract $.5$ from it to get the probability for $[0, t]$. If your cumulative probability is less than $.5$, subtract your cumulative probability from $.5$ and negate to get the probability of falling into $[t, 0]$. Since the normal curve is symmetric, this is the same as the probability of falling into $[0, t]$.


As for using "slope intercept" to find the exact value, it sounds like your instructor just wants you to interpolate values that aren't on the table. For example, if you get a probability of $.2$, that's $x$ portion of the way between  the probabilities $.196$ and $.226$ corresponding to $t = .5$ and $t = .6$ on your table. Hence, you want a value of $t$ that is $x$ portion of the way between $.5$ and $.6$. All that's left is to find $x$:
$$x = \frac{.2 - .192}{.226 - .192} = \frac{.004}{.034} = .1176 \ldots$$
Then you have your $z$-score: $$t = .5 + x * (.6 - .5) = .5 + .0117 \ldots = .5117 \ldots $$
