I am trying to model a of (normal-distribution-like) discrete random variable using the normal distribution.
This is what I understand so far:
First, I approximate the mean of the normal distribution using the mean of the observed values, i.e.
$\mu = \frac{x_1 + x_2 + ... + x_n}{n}$
Then I approximate the standard deviation:
$\sigma = \sqrt{\frac{(x_1 - \mu)^2 + (x_2 - \mu)^2 + ... + (x_n - \mu)^2}{n}}$
Then its PDF would be
$P(X \lt x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x - \mu)^2}{2\sigma^2}}$
Then since what I have is basically discrete random variables, the probability of observing a discrete value (e.g. $2$) can be approximated by
$P(1.5 \lt X \lt 2.5) = P(X \lt 2.5) - P(X \lt 1.5)$
Are the above OK so far?
Another thing is, from the PDF, $P(X \lt 0)$ is a non-zero value. However, the nature of my random variable is that $0$ is the minimum that can occur. How can I model the random variable, such that it does follow normal distribution for $X \ge 0$, but that $P(X \lt 0) = 0$? (Note that there is no cut off at the upper end, i.e. the upper tail can stretch infinitely).
You can use truncated normal distribution as you describe, but your procedure for estimating its parameters ($\mu$, $\sigma$) is not valid. Try deriving the maximum likelihood estimator.
Alternatively, you may try a proper discrete distribution such as binomial distribution which has a similar shape.