# Computation of parameters of truncated normal distribution

I have a numeric data sample with mean $\mu$ and standard deviation $\sigma$. I believe that this data has a normal distribution truncated to $[0,1]$.

1. Is there a reasonably simple formula for estimating the parameters $\mu', \sigma'$ of that truncated normal distribution? Is there an R command which does it?

2. I understand that I could find $\mu', \sigma'$ by applying a maximum-likelihood fitting algorithm directly to my data (through, say, fitdistr in R).

3. Is there some good understanding how much more precise approach (2) is over approach (1)? Say the sample size is from 10 to 100, $\mu=0.75,$ $\sigma=0.15$.

• Out of curiosity--because this is a pretty unusual thing to encounter--what application is it that models data in $[0,1]$ using truncated normal distributions? – whuber Jun 27 '14 at 19:07
• @whuber: I chose truncated normal distribution to model student percentile scores. I made this decision based on available data to me. There is no strong theoretical reason to choose it, but (as far as I know) neither there is one for other distributions. – Adam Jun 27 '14 at 19:47
• There are some strong arguments for alternatives, especially censored distributions: this is what happens when lots of people achieve the top score (or near enough to it, up to measurement error). The scores can be viewed as censored versions of an instrument that in principle allows even higher scores, but all scores over 100% are just recorded as 100%. (I don't know if the College Board still does this, but at least once upon a time they internally graded SATs on a scale up to 900, but reported all grades in the 800-900 interval as 800: that's the same idea.) – whuber Jun 27 '14 at 21:42
• I'm voting to close this question as off-topic because it seems to be abandoned by the OP and otherwise to localized. – kjetil b halvorsen Apr 21 '17 at 21:05
• Thanks, @whuber. That sounds interesting! My data are student percentage scores for a course. I noticed that when one eliminates the weak scores (say below 55%) then what's left fits a truncated normal distribution very well and my question is about finding its (untruncated) mean and variance. – Adam Apr 26 '17 at 0:13