# 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:

$$\begin{array}{|} \hline \text{\quad\quadTabled values of the normal curve}\\ \hline \text{area}\ 0-t & t\\ 0 & 0\\ 0.39 & 0.1 \\ 0.79 &0.2 \\ 0.118 & 0.3\\ 0.155 & 0.4\\ 0.192 & 0.5\\ 0.226 & 0.6\\ 0.258 & 0.7\\ 0.288 & 0.8\\ 0.316 & 0.9\\ 0.341 & 1.0\\ 0.364 & 1.1\\ 0.385 & 1.2\\ 0.403 & 1.3\\ 0.419 & 1.4\\ 0.433 & 1.5\\ 0.445 & 1.6\\ 0.455 & 1.7\\ 0.464 & 1.8\\ 0.471 & 1.9\\ 0.477 & 2.0\\ 0.482 & 2.1\\ 0.486 & 2.2\\ 0.489 & 2.3\\ 0.492 & 2.4\\ 0.494 & 2.5\\ \hline \end{array}$$

• nice ................+1 @RockOn. Feb 5, 2016 at 18:20
• What kind of distribution is this? That table does not look like what I expected (normal) Feb 5, 2016 at 18:25
• @StellaBiderman The only other mentioning of it states: "Use attached table for unit normal deviates find nearest 0-t entry and use exact tabled value without rounding or interpolation." Does that mean anything to you? Feb 5, 2016 at 18:28

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: $$t = .5 + x * (.6 - .5) = .5 + .0117 \ldots = .5117 \ldots$$

• Wow thank you so much for the explanation. It is much clearer to me. Now I am wondering, how can I look up a probability like 0.7308 when the provided table doesn't go beyond 0.494? Feb 5, 2016 at 19:12
• @RockOn Fixed my answer. It looks like you actually are calculating cumulative probabilities (i.e., the usual way of doing z-scores), since otherwise $0.7308$ wouldn't make sense (the area under the entire right half of the curve is only $.5$, since the normal curve is symmetric and has total area 1.) Feb 5, 2016 at 19:14
• Ah okay! I see. Thank you!! Feb 5, 2016 at 19:23
• (Thanks for adding the extra pics as well. You really helped me a lot with this.) Feb 5, 2016 at 19:34
• This is excellent and the problem is awful. Feb 5, 2016 at 20:35