I'm working on the Boston housing project of the udacity ML nano degree. A histogram of the data set looks like this:

enter image description here

The mean, median and mode are:

mean:   454342.944785
median: 438900.0
mode:   525000

Is it correct to say that it has a normal distribution?

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    $\begingroup$ for two probability distribution to be exactly the same, they need to have the same moments for all the orders up to infinity. mean is order 1, variance is 2. skewness is 3 etc. So just comparing 2 or 3 measures is not enough. $\endgroup$ Aug 4, 2017 at 17:39
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    $\begingroup$ Consider carefully why you care whether the data themselves have a normal distribution (as the answers below show they cannot, in any event). What can be important in modeling is whether the residuals are normally distributed, as that assumption underlies many standard statistical tests. Even then, some tests are fairly insensitive to departures of residuals from normality, and approaches like bootstrapping can provide useful tests and confidence intervals without normality assumptions. $\endgroup$
    – EdM
    Aug 4, 2017 at 17:50
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    $\begingroup$ Whether the distribution is normal enough for using some statistical test is another matter. stats.stackexchange.com/questions/2492/… $\endgroup$ Aug 5, 2017 at 1:49

2 Answers 2


How far can be median, mode and mean to still say that is a normal distribution?

This kind of gets things backward -- even if they were exactly equal, that's no basis on which to claim you have a normal distribution. Note that

  1. the population values are equal for many distributions that are not normal (if the population values were unequal of course you'd have non-normality, but if they're all equal it doesn't tell us that you have symmetry)

  2. the sample values could be equal or very close to it even if the population values differ (indeed exact equality would suggest the distribution was discrete, and therefore not normal).

If you're using the data I think you are, for the variable you're referring to you have discretized and censored data, so normality would be moot. We can also see that it can't be normal because house values can't be negative.

So one thing you can say with confidence is that those values you have are not drawn from a normal distribution

Leaving that specific data aside, what we can do instead of trying to day data come from a normal distribution when those location values are close together is to ask "how far apart would they have to be to say that they're inconsistent with normality?".

That we can do something with, at least with respect to mean and median. (The sample mode is a bit tricky with continuous distributions; it would depend on how you obtain it; I suggest we leave that issue aside.)

The distance that sample mean and median would tend to differ will depend on scale and sample size. So one way to assess that difference independent of scale would be to measure how many standard deviations they are apart.

Note that (mean-median)/s.d. is one third of the second Pearson skewness; it's also (apparently) sometimes called the nonparametric skew.

So let's define that statistic, $$S=\frac{\bar{x}-\tilde{x}}{s}$$ (where $\tilde{x}$ is the sample median), which is one on which we can base a test.

Doane & Seward (2011)[1] offer a brief table for a test of $3S$ (the second Pearson skewness) at the normal.

Cabilio and Masaro (1996)[2] use $S$ as a test statistic for a test of symmetry (based on the values at the normal).

[In their case the test is asymptotic; you'd reject symmetry if $|S|>0.7555 \,Z_{\alpha/2}/\sqrt{n}$. Simulations suggest the asymptotic values aren't too bad once you get some way beyond the ends of Doane and Seward's table, I'd consider using it upward from about $n=400$ or so, though there's only about two figure accuracy in the critical values.]

Note that using this sort of statistic to decide if your distribution is non-normal would leave you unable to reject many other distributions (including -- in spite of Cabilio & Masaro's test being for asymmetry -- some asymmetric distributions which have mean = median)

[1]: Doane, D. P., Seward L. E. (2011),
Measuring Skewness: A Forgotten Statistic?
Journal of Statistics Education, Volume 19, Number 2

[2]: Cabilio, P. & Masaro, J. (1996),
A Simple Test of Symmetry about an Unknown Median,
The Canadian Journal of Statistics, Vol. 24, No. 3 (Sep.), pp. 349-361


No the distribution is certainly not normal. Normal distribution has a bell-curve shape and has mean = mode = median. However normal distribution is a synthetic mathematical function and no real life data will look exactly normal (even if you use random number generator to get draws from the normal distribution, even relatively small samples may not "look" normal and pass normality tests). In real-life we are interested is approximate normality, it is your decision if you want to use normal distribution as an approximation for the distribution of your data. Usually to convince yourself if the distribution behaves well, you would look at density plots, Q-Q plots, Shapiro-Wilk tests, Kolmogorov-Smirnov tests etc., but as describes in the Is normality testing 'essentially useless'? thread, those methods may misguide you and there is no fool-proof method to decide that the data "is", or "is not" normal. Moreover, in real life we often use normal approximation for data that is certainly not-normal, e.g. for survey data on Likert-type scales (bounded, discrete).

all models are wrong, but some are useful - G. E. P. Box


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