The definition for the probability density function is:
p[x_, variance_] := 2/Sqrt[π variance^2] Exp[-x^2/variance^2]
Here is a plot with different variance parameters.
Plot[{p[x, 0.2], p[x, 0.4], p[x, 0.6], p[x, 0.8], p[x, 1.0]},
{x, -2, 2}, PlotRange -> All]

Here is the definition of the cumulative distribution function. We force the input arguments to be numerical to facilitate subsequent processing in FindRoot
(avoids error messages).
cdf[variance_ /; variance > 0, limit_ /; limit > 0] :=
NIntegrate[p[x, variance], {x, 0, limit}]
We are interested in the correlation between the variance and the x
value corresponding to cdf = 0.5
.
Here is a graph of it with different variance parameters.
Show[
Plot[{cdf[0.2, x], cdf[0.4, x], cdf[0.6, x], cdf[0.8, x],
cdf[1.0, x]}, {x, 0, 1}, PlotRange -> {0, 1}],
Plot[0.5, {x, 0, 2}, PlotStyle -> Black]
]

Here is a table of variance values and the corresponding x
value where the cdf
is 0.5.
median[variance_] := FindRoot[cdf[variance, limit] == 0.5, {limit, 0.2}]
varianceData =
Table[{limit /. median[variance], variance}, {variance, 0.2, 1, 0.2}]
(* {{0.0953873, 0.2}, {0.190775, 0.4}, {0.286162,
0.6}, {0.381549, 0.8}, {0.476936, 1.}} *)
We superimpose it on our plot.
Show[
Plot[{cdf[0.2, x], cdf[0.4, x], cdf[0.6, x], cdf[0.8, x],
cdf[1.0, x]}, {x, 0, 1}, PlotRange -> {0, 1}],
Plot[0.5, {x, 0, 2}, PlotStyle -> Black],
ListPlot[
Transpose@Join[{varianceData[[All, 1]]}, {ConstantArray[0.5, 5]}],
PlotStyle -> {PointSize[0.025], Red}]
]

The question is, given a particular x = α
, what is the corresponding variance?
That can be solved using the same procedure as median
but swapping the variable and parameter.
fun[α_] := FindRoot[cdf[variance, α] == 0.5, {variance, 0.2}]
Let's test it by using as inputs the limit
value from varianceData
and see if we get back the same variances.
limitData =
Table[{limit, variance /. fun[limit]},
{limit, varianceData[[All, 1]]}]
(* {{0.0953873, 0.2}, {0.190775, 0.4}, {0.286162,
0.6}, {0.381549, -0.8}, {0.476936, -1.}} *)
Sure enough, seems to work fine.
p[x]
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