While the article you refer to correctly defines the concept of confidence interval (your highlighted text) it does not correctly treat the case of a normal distribution with unknown standard deviation. You may want to search "Neyman confidence interval" to see an approach that produces confidence intervals with the property you highlighted.
The Neyman procedure selects a region containing 95% of outcomes, for each true value of the parameter of interest. The confidence interval is the union of all parameter values for which the observation is within the selected region. The probability for the observation to be within the selected region for the true parameter value is 95%, and only for those observations, will the confidence interval contain the true value. Therefore the procedure guarantees the property you highlight.
If the standard deviation is known and not a function of the mean, the Neyman central confidence intervals turn out to be identical to those described in the article.
Thank you for the link to Neyman's book - interesting to read from the original source! You ask for a simple description, but that is what my second paragraph was meant to be. Perhaps a few examples will help illustrate: Example 1 and 1b could be considered trivial, whereas 2 would not be handled correctly by the article you refer to.
Example 1. Uniform random variable. Let X follow a uniform distribution,
$$f(x)=1/2 {\mathrm{\ \ for\ \ }}\theta-1\le x\le \theta+1 $$ and zero otherwise.
We can make a 100% confidence interval for $\theta$ by considering all possible outcomes $x$, given $\theta$, ie. $x \in [\theta-1,\theta+1]$. Now consider an observed value, $x_0$. The union of all possible values of $\theta$ for which $x_0$ is a possible outcome is $[x_0-1,x_0+1]$. That is the 100% confidence interval for $\theta$ for this problem.
Example 1b. Uniform random variable. Let X follow the same uniform distribution. We can make a 95% central confidence interval for $\theta$ by selecting the 95% central outcomes $x$, given $\theta$, ie. $x \in [\theta-0.95,\theta+0.95]$. Now consider an observed value, $x_0$. The union of all possible values of $\theta$ for which $x_0$ is within the selected range is $[x_0-0.95,x_0+0.95]$. That is the 95% confidence interval for $\theta$ for this problem.
Example 2. Uniform random variable. Let X follow a uniform distribution,
$$f(x)=1/\theta {\mathrm{\ \ for\ \ }}{1\over2}\theta \le x \le {3\over2}\theta $$ and zero otherwise. We can make a 100% confidence interval for $\theta$ by considering all possible outcomes $x$, given $\theta$, ie. $x \in [{1\over2}\theta,{3\over2}\theta]$. Now consider an observed value, $x_0$. The union of all possible values of $\theta$ for which $x_0$ is a possible outcome is $[{2\over3}x_0,2x_0]$. That is the 100% confidence interval for $\theta$ for this problem. (You can confirm this by inserting the endpoints of the confidence interval into the pdf and see they are at the boundaries of the pdf). Note that the central confidence interval is not centered on the point estimate for $\theta$, $\hat\theta = x_0$.
Example 3. Normal distribution with mean $\theta$ and standard deviation $1$. The 68% central confidence interval would be constructed identically to example 1, that is the selected region for $X$ would be $[\theta-1,\theta+1]$. The 68% central confidence interval is therefore the same as in Example 1, $[x_0-1,x_0+1]$. You can extend this to 95% and arbitrary KNOWN standard deviation $\sigma$ to be $[x_0-1.96\sigma,x_0+1.96\sigma]$.
Example 4. Normal distribution with mean $\theta$ and standard deviation $\theta/2$. The 68% central confidence interval would be constructed identically to example 2. The 68% central confidence interval for $\theta$ is therefore the same as in Example 2, $[{2\over3}x_0,2x_0]$.
The authors of the article you refer to and the other commenters to your question would not get Example 2 or 4 right. Only following a procedure like Neyman's will the confidence interval have the property that you highlighted in your post. The other methods are approximations for the general problem of building confidence intervals.
The exact solution to the problem with a normal distribution and UNKNOWN standard deviation is more difficult to work out than the examples above.
There is a simple, well-known modification to skew a normal distribution as follows,
$$Y_k(X) = Y(X) + k\frac{dY(X)}{dX}$$
where $k$ is the skew parameter to regulate the amount of skews and
$$\frac{dY(X)}{dX}= \frac{1}{\sqrt{2\pi S^2} } e^{ -\frac{(X-A)^2}{(2S)^2} }\left(-\frac{X-A}{2S^2}\right)
$$
This is often used in financial modeling, and it preserves the unit cumulative integration.
Best Answer
You can use the skew normal distribution with parameters $(ξ,ω,α)$ which can be estimated from the given data. If we set $δ=\dfrac{α}{\sqrt{1+α^2}}$, then the mean, variance and skewness of the skew normal distribution are given by (see the link)
Substitute your known values for the mean, variance and skewness to find proper values for the parameters $(ξ,ω,α)$ of the distribution. Approximate values will do (you do not need to solve exactly), because you are based on a sample, and the distribution that you will find does not need to fit exactly to the sample. So, trial and error (with a computer), may help, since this is not an easy to solve system.
Start from the formula of the skewness which depends only on $δ$. That is solve
$$0.3131=\dfrac{4-\pi}{2} \dfrac{\left(\delta\sqrt{\dfrac2\pi}\right)^3}{ \left(1-\dfrac{2\delta^2}{\pi}\right)^{3/2}}$$ to find (approximately $δ$). From $δ$ you can find directly $α$. Now, go to the variance and solve $$0.028=ω^2\left(1-\dfrac{2δ^2}{π}\right)$$ to find $ω$. Use the value for the $δ$ that you already have. Finally, use the formula for the mean (with $ω, δ$ known) to find $ξ$ in a similar way.