Suppose I draw some values from $X_i \sim {\mathcal{N}( \mu, \, \sigma^2 )} $ and then round them to the nearest integer. Are those rounded values then Binomial distributed?

  • $\begingroup$ no, because binomial distribution is valued on a finite number of integers $\endgroup$
    – lion
    Nov 8, 2017 at 14:16
  • $\begingroup$ But is it a reasonable approximation? $\endgroup$
    – PascalIv
    Nov 8, 2017 at 14:40

1 Answer 1


Rounding to integers can seriously degrade the data for some purposes, especially if the population variance is small. The effect of rounding on sample means and variances is often minimal, so t methods may not be seriously affected. You say the original data are normal, so you would not likely use rank-based tests, but they are seriously affected by rounding because of the potential for creating ties.

Analyses on paired data can be especially sensitive to rounding, even using t methods. If differences are small and have small variances, you can 'round away' a significant difference by rounding to integers (instead of say tenths). One way to check the damage from rounding is to look at the proportion of ties in the rounded data.

Only in very unusual circumstances would rounded normal data be anything like binomial data which take integer values from $0$ up to some $n.$ (Try to imagine the binomial model, what would be a 'trial', what would be a 'success'?)

Here is a quick experiment. I generate 100 observations from $\mathsf{Norm}(\mu=20,\, \sigma=2)$ and round to integers, then check for ties. The original data are generated to many decimal places and ties are extremely rare (none in my experiment). But after rounding to integers, there remain only ten uniquely different values among the 100 observations. However, sample means and SDs are not much changed by rounding to integers. (See the R output below.) By contrast, rounding to two places leaves 91 uniquely different values (not shown).

x = rnorm(100, 20, 2);  length(unique(x))
## 100  # 100 uniquely different observations
rx = round(x);  length(unique(rx))
## 10   # only 10 uniquely different obs after rounding
mean(x); sd(x)
## 19.95831    # sample mean before rounding
## 2.084899    # sample SD before
mean(rx); sd(rx)
## 19.94       # sample Mean after rounding
@@ 2.088061    # sample SD after

Note: Predicting ties in rounded data is somewhat like the famous birthday problem, except that different values have different probabilities of producing ties ('matches'). In the case with $\mu = 20$ and $\sigma = 2,$ one might reasonably expect most values to be in the interval $(14, 26).$ That leaves only 13 integer values to which one might round.

  • $\begingroup$ Is it mathematically correct to write this "rounded" distribution as: $ X_i \sim \left \lfloor {\mathcal{N}( \mu, \, \sigma^2 )} \right \rceil$ ? $\endgroup$
    – PascalIv
    Nov 9, 2017 at 14:34
  • $\begingroup$ I know of no standard notation for normal data rounded to integers. Yours seems fine, but would have to be explained. Until your method became standard, I would say "$N(\mu,\sigma^2)$ rounded to integers." $\endgroup$
    – BruceET
    Nov 9, 2017 at 15:49

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