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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}

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  • $\begingroup$ nice ................+1 @RockOn. $\endgroup$ Feb 5, 2016 at 18:20
  • $\begingroup$ What kind of distribution is this? That table does not look like what I expected (normal) $\endgroup$ Feb 5, 2016 at 18:25
  • $\begingroup$ @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? $\endgroup$
    – RockOn
    Feb 5, 2016 at 18:28

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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: Illustration of your table, in beautiful MS paint form

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 $[0, t]$.

Does this belong in a modern art museum? You decide. 508 compliance? What's 508 compliance? This is the Internet!

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

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  • $\begingroup$ 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? $\endgroup$
    – RockOn
    Feb 5, 2016 at 19:12
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    $\begingroup$ @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.) $\endgroup$ Feb 5, 2016 at 19:14
  • $\begingroup$ Ah okay! I see. Thank you!! $\endgroup$
    – RockOn
    Feb 5, 2016 at 19:23
  • $\begingroup$ (Thanks for adding the extra pics as well. You really helped me a lot with this.) $\endgroup$
    – RockOn
    Feb 5, 2016 at 19:34
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    $\begingroup$ This is excellent and the problem is awful. $\endgroup$ Feb 5, 2016 at 20:35

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