<p><a href="https://en.wikipedia.org/wiki/Sensitivity_index" rel="nofollow noreferrer">d'</a> is equal to |mean<sub>A</sub> - mean<sub>B</sub>|/SD when you have two gaussian distributions of equal variance and different means.</p>
<p>Imagine SD=1, mean<sub>A</sub> = 0, mean<sub>B</sub> = 1, with B&gt;A as the &quot;correct&quot; answer (perhaps B is the signal, and A is the noise). This will give you a d' of 1 because the difference between the means is 1, divided by a standard deviation of 1.</p>
<p>The % correct would be the fraction of the time that a random draw from the distribution with larger mean is larger than a draw from the distribution with smaller mean. That's what an ideal observer would do given the task to choose the larger distribution: pick the larger number on that trial. This is equivalent to detecting a signal if you think of the other distribution as representing the &quot;noise&quot;.</p>
<p>For this example (or any example that satisfies the condition that d' = 1), that rate will be about 76%. I don't believe there is a simple closed form equation to calculate this, and my calculus class days for rearranging integrals are a bit dated, but you can simulate it easily by taking (pseudocode):</p>
<pre><code>a = rand(mu=0,sigma=1,n=100000)
b = rand(mu=1,sigma=1,n=100000)
sum(b&gt;a)/100000
</code></pre>
<p>Note that there isn't any special reason d' = 1 is chosen to make this 76% number be true, it's just a consequence of the assumption of normal distributions and probabilities. If d' = 0, the % &quot;correct&quot; will be 50%.</p>

CC BY-SA 4.0
2021-12-22T11:41:37.400Z
Thanks @Bryan, I understand this, but my question is **Why 76%**? I've never seen a satisfactory derivation of this figure. I am specifically interested in where the 76% figure comes from.
CaptainProg
CC BY-SA 4.0
2021-12-22T15:12:40.597Z
@CaptainProg Uh, I've explained it here, no? Draw from two normal distributions; if d' for those distributions = 1, then the one that is larger on average will be larger in your draw 76% of the time. I don't believe there is a simple closed form equation to calculate this, and my calculus class days for rearranging integrals are a bit dated, but you can simulate it easily by taking (pseudocode) a = rand(mu=0,sigma=1), b= rand(mu=1,sigma=1) a bunch of times and checking how often b>a
Bryan Krause
CC BY-SA 4.0
2021-12-22T15:36:32.740Z
Thanks for editing – yes, this now makes sense. I've run this in MATLAB and finally happy that I can see how 76% was derived (with 100,000 random samples generated, I got 75.99%)
CaptainProg
CC BY-SA 4.0
2021-12-21T21:01:43.183Z
2021-12-22T15:14:35.377Z
2021-12-22T15:14:35.377Z
Bryan Krause
<p>In signal detection theory using a 2-alternative forced choice task, when d' = 1, threshold is found at the stimulus intensity which is perceived 76% of the time (<a href="https://www.oxfordreference.com/view/10.1093/oi/authority.20110803110358463" rel="nofollow noreferrer">https://www.oxfordreference.com/view/10.1093/oi/authority.20110803110358463</a>).</p>
<p>Why 76%?</p>

<psychophysics><signal-detection>
Why is threshold at 76% correct when d-prime = 1?
CaptainProg
# Why is threshold at 76% correct when d-prime = 1?

(Asked by: username_0 on 2021-12-21)

In signal detection theory using a 2-alternative forced choice task, when d' = 1, threshold is found at the stimulus intensity which is perceived 76% of the time (<https://www.oxfordreference.com/view/10.1093/oi/authority.20110803110358463>).

Why 76%?

**Tags:** psychophysics, signal-detection


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## 1 answers

### Answer by username_1 on 2021-12-21. Score: 2 | Accepted answer

d' is equal to |meanA - meanB|/SD when you have two gaussian distributions of equal variance and different means.
Imagine SD=1, meanA = 0, meanB = 1, with B>A as the "correct" answer (perhaps B is the signal, and A is the noise). This will give you a d' of 1 because the difference between the means is 1, divided by a standard deviation of 1.
The % correct would be the fraction of the time that a random draw from the distribution with larger mean is larger than a draw from the distribution with smaller mean. That's what an ideal observer would do given the task to choose the larger distribution: pick the larger number on that trial. This is equivalent to detecting a signal if you think of the other distribution as representing the "noise".
For this example (or any example that satisfies the condition that d' = 1), that rate will be about 76%. I don't believe there is a simple closed form equation to calculate this, and my calculus class days for rearranging integrals are a bit dated, but you can simulate it easily by taking (pseudocode):
a = rand(mu=0,sigma=1,n=100000)
b = rand(mu=1,sigma=1,n=100000)
sum(b>a)/100000

Note that there isn't any special reason d' = 1 is chosen to make this 76% number be true, it's just a consequence of the assumption of normal distributions and probabilities. If d' = 0, the % "correct" will be 50%.

> **Comments:**
> **username_0**: Thanks @Bryan, I understand this, but my question is **Why 76%**? I've never seen a satisfactory derivation of this figure. I am specifically interested in where the 76% figure comes from.  
> **username_1**: @username_0 Uh, I've explained it here, no? Draw from two normal distributions; if d' for those distributions = 1, then the one that is larger on average will be larger in your draw 76% of the time. I don't believe there is a simple closed form equation to calculate this, and my calculus class days for rearranging integrals are a bit dated, but you can simulate it easily by taking (pseudocode) a = rand(mu=0,sigma=1), b= rand(mu=1,sigma=1) a bunch of times and checking how often b>a  
> **username_0**: Thanks for editing – yes, this now makes sense. I've run this in MATLAB and finally happy that I can see how 76% was derived (with 100,000 random samples generated, I got 75.99%)  

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