You seem to have drawn a density function. $\mu$ and $\sigma^2$ do not need to be the mean and variance of your values to do this.
So try to draw a second similar curve for the function with the same $v_i$ and slightly different values for $\mu$ and $\sigma^2$. If the two $\mu$s are close enough (depending on the $\sigma^2$s) then there should be a visible overlap. Any value $v$ appearing in the middle of this overlap could reasonbly come from either distribution.
Strictly speaking, a probability distribution is a function (more precisely, a measure) that assigns to each event some real number in $[0,1]$. Whenever $X$ is a random variable, giving its probability distribution is giving the probabilities attached to the values that $X$ can take. For example if $X$ is the number given when you roll a die, and if $P_X$ is its probability distribution, then you have $P_X(\{1,2\})={1 \over 3}$, $P_X(\{3\})={1 \over 6}$ and so on. For each event, you assign a real number that is the probability of this event. This function is called the probability distribution of $X$.
Now probability distribution is also used is a broader sense, which is closer to the meaning of statistical model. For example, we say $X$ has the binomial distribution. When we say that, what we really mean is $X$ has a binomial distribution, that is: there exists some $n$ and $p$ such that $X\sim Bin(n,p)$. But strictly speaking, if $X\sim Bin(3,0.2)$ and $Y\sim Bin(3,0.4)$, $X$ and $Y$ don't have the same distribution because the probabilities are not the same. However we talk about the binomial distribution.
That is where the concept of statistical model arises. A statistical model is just a set of probability distributions $\mathcal{P}=\{P_{\theta},\theta\in\Theta\}$. For example $\Theta=(0,1)$ and $P_\theta$ is the $Bin(10,\theta)$ distribution. This is a binomial model. This is used in statistics, when you have observations, but you don't know the underlying probability distribution. Thus we make the hypothesis that it belongs to some set of distributions, which is your model, indexed by a parameter $\theta$. Then, we ask ourselves: based on the observations, what can we say about $\theta$, what is the real underlying distribution? Note that we may be wrong. Maybe the real distribution does not belong to this statistical model.
Short Answer
- A probability distribution is a function that assigns to each event a number in $[0,1]$ which is the probability that this event occurs.
- A statistical model is a set of probability distributions. We assume that the observations are generated from one of these distributions.
Best Answer
It sounds to me like your primary difficulty is in binning the data properly. Please pardon me if this is not your question.
You want the total area of the rectangles comprising the histogram to be 1. If you take rectangle height to be the number of data points in the corresponding bin, then the total area will not be 1. Instead, you will have $$ \text{area}=(\text{bin width})\times(\text{total number of data points)}. $$ The general principle for handling distributions of binned data properly - which even allows you to deal with variable-width bins - is that rectangle area, rather than rectangle height, should be proportional to the number of data points in the bin. For the fixed-width case, the height of each rectangle should be computed as $$ \text{height}= \frac{(\text{number of data points in bin})}{(\text{total number of data points})\times(\text{width of bin})}. $$
If you follow this procedure, then, within reason, the resulting distribution should be relatively insensitive to the number of bins you decide to use. Of course, if you use too many bins, most bins will be empty and you'll get a very spiky distribution, which won't be very illuminating. If you use too few bins, the distribution will be too coarse-grained. The Wikipedia page on histograms describes some commonly used rules, such as Sturges' formula, for deciding how many bins to use.
For testing goodness of fit, you should follow the suggestions in Michael Chernick's answer.