Calculate probability (area) under the overlapping area of two normal distributions I have two normal distributions defined by their averages and standard deviations.
Sample 1: Mean=5.28; SD=0.91
Sample 2: Mean=8.45; SD=1.36
You can see how they look like in the next image:

How can I get the probability to obtain an individual from the overlapping area (green)? Is the probability the same as the area?
 A: There are two scenarios when calculating the intersections of two normal distributions. It's important to acknowledge that, in most cases, there are two intersection points:
Equal variance
In the modelled (or rare) examples when two normal distribution curves have exactly the same variance, there is one intersection point.  The intersection point on the abscissa axis, $(x_1)$, is calculated from:
$$x_1=\frac{\mu_a+\mu_b}{2}$$
This being, of course, the midpoint between the respective means.
Unequal Variance
When the two normal distribution curves intersect and have different variances, the solution to the intersection points simplifies to a quadratic form.  The consequence of the quadratic form is that there are always two intersection points $(x_1,x_2)$ on the abscissa axis.
The intersection points of the two curves can be solved algebraically to obtain the quadratic form and the roots can then be found using the quadratic formula. It’s not as hard as it seems and does not involve using the error function. However, for time and space, I give Inman and Bradley’s version where the intersection points can be found directly.
$$(x_1,x_2)=\frac{\mu_a \sigma_b^2-\mu_b \sigma_a^2\pm\sigma_a \sigma_b \sqrt{(\mu_a-\mu_b)^2+(\sigma_b^2-\sigma_a^2)\ \ln \left(\frac{\sigma_b^2}{\sigma_a^2}\right)}}{\sigma_b^2-\sigma_a^2}$$
Where $-\infty<{x_1}<{x_2}<\infty$. (For ease of calculation, I set $\sigma_b>\sigma_a$ ).
For some intersecting normal distributions with unequal variances, the ‘second’ intersection point maybe insignificant and can, effectively, be ignored. However, for most cases – especially when $\mu_a\approx\mu_b$ – having both intersection points are critical to determine the ‘common area’ (overlap).
Finding the Area
The common area (overlap), $\phi$, is calculated from:
Equal variance
$$\phi=\min\left[\int_{-\infty}^{x_1} f(a) dx, \int_{-\infty}^{x_1} f(b) dx\right]+\min\left[\int_{x_1}^{\infty} f(a) dx, \int_{x_1}^{\infty} f(b) dx\right]$$
Where, of course, ${f(a)}=\frac{1}{\sigma_a \sqrt{2 \pi}} e^{-0.5 \left(\frac{x-\mu_a}{σ_a}\right)^2}$
Unequal variance
$$\phi=\min\left[\int_{-\infty}^{x_1} f(a) dx, \int_{-\infty}^{x_1} f(b) dx\right]+\min\left[\int_{x_1}^{x_2} f(a) dx, \int_{x_1}^{x_2} f(b) dx\right]+\min\left[\int_{x_2}^{\infty} f(a) dx, \int_{x_2}^{\infty} f(b) dx\right]$$
Answer
Inserting the data from the question, $\mu_a=5.28, \sigma_a=0.91$ and $\mu_b=8.45, \sigma_b=1.36$, we get $x_1=-1.2842, x_2=6.7046$ and $\phi=0.1548$
Reference:
Inman, H.F. & Bradley Jr, E.L. (1989). The overlapping coefficient as a measure of agreement between probability distributions and point estimation of the overlap of two normal densities, Communications in Statistics - Theory and Methods, 18:10, 3851-3874, DOI: 10.1080/03610928908830127
A: It is not quite clear what you mean by probability to obtain an individual from the overlapping area. This solves for the area of the green zone in your diagram:
Let:


*

*$X_1 \sim N(\mu_1,\sigma_1^2)$ with pdf $f_1(x_1)$ and cdf $ F_1(x_1)$ and 

*$X_2 \sim N(\mu_2,\sigma_2^2)$ with pdf $f_2(x_2)$ and cdf $ F_2(x_2)$, 
where $\mu_1 < \mu_2$. In your example, the 'black variable' corresponds to $X_1$.
Let $c$ denote the point of intersection where the pdf's meet in the green zone of your plot Then, the area of your green intersection zone is simply:
$$P(X_1>c) + P(X_2<c) = 1 - F_1(c) + F_2(c) = 1-\frac{1}{2} \text{erf}\left(\frac{c-\mu _1}{\sqrt{2} \sigma _1}\right)+\frac{1}{2} \text{erf}\left(\frac{c-\mu _2}{\sqrt{2} \sigma _2}\right)$$
where erf(.) is the error function.
Point $c$ is the solution to $f_1(x) = f_2(x)$ within the green zone, which yields:
$$c =    \frac{\mu _2 \sigma _1^2-\sigma _2 \left(\mu _1 \sigma _2+\sigma _1 \sqrt{\left(\mu _1-\mu _2\right){}^2+2 \left(\sigma _1^2-\sigma _2^2\right) \log \left(\frac{\sigma _1}{\sigma _2}\right)}\right)}{\sigma _1^2-\sigma _2^2}$$
For your example, with $ {\mu_1 = 5.28, \mu_2 = 8.45, 
  \sigma_1 = 0.91, \sigma_2 = 1.36}$, this yields:  $c = 6.70458...$,
and the area of the green section is:  0.158413 ...

A: @abdelbasset,
To improve @wolfies 's answer above, there are two intersection points c, let's call them c1 and c2. Here, c1=-1.2848 and c2=6.7046. @wolfies ignored c1 (it likely will be too far in the tails of the probability density functions to matter (especially if we round to just a few decimal points)).

The more correct way is to find both c1 and c2, and to find the area of overlap of both functions: 
AREA OF OVERLAP = P(X1 > j1) + P(X2 < j1) - [P(X2 < j2)-P(X1 < j2)]
therefore, using cdf's
AREA OF OVERLAP = 1 - F1(j1,μ2,σ2)+F2(j1,μ2,σ2)-F2(j2,μ2,σ2)+F1(j2,μ1,σ1)
A: Normalise the graphs to an area of 1 by dividing each by their respective standard deviation. Then use simple subtraction from a z-graph to calculate the probability of an occurrance in that overlap area. No need for erfs.
