I created a histogram for Respondent Age and managed to get a very nice bell-shaped curve, from which I concluded that the distribution is normal.

Then I ran the normality test in SPSS, with n = 169. The p-value (Sig.) of the Kolmogorov-Smirnov test is less than 0.05 and so the data have violated the assumption of normality.

Why does the test indicate that the age distribution is not normal, but the histogram showed bell-shaped curve, which from my understanding is normal? Which result should I follow?

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    $\begingroup$ Why are you testing normality? $\endgroup$
    – Glen_b
    Commented Dec 17, 2014 at 5:05
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    $\begingroup$ In addition to @Glen_b's excellent comment and Aksakal's equally excellent answer, note that even for continuous distributions, K-S requires that the mean and sd be known beforehand, not estimated from the data. This essentially makes the K-S test useless. "The Kolmogorov-Smirnov test is only a historical curiosity. It should never be used." (D'Agostino in d'Agostino & Stephens, eds., 1986). If at all, use Shapiro-Wilks instead. $\endgroup$ Commented Dec 17, 2014 at 8:18
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    $\begingroup$ @Stephan Kolassa Good advice, but you mean Shapiro-Wilk. (The suggestions of M.B. Wilk and S.S. Wilks are often confused or conflated; the strange use of 's as possessive in English here also can contribute to the confusion, even for many who have English as their first language.) $\endgroup$
    – Nick Cox
    Commented Dec 17, 2014 at 10:46
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    $\begingroup$ Related to the comment of @StephanKolassa, see Is Shapiro-Wilk the best normality test? ... the answer is that it isn't necessarily, depending on which alternative you're interested in, but it is very often a good choice. $\endgroup$
    – Silverfish
    Commented Jan 5, 2015 at 22:14

2 Answers 2


We usually know it's impossible for a variable to be exactly normally distributed...

The normal distribution has infinitely long tails extending out in either direction - it is unlikely for data to lie far out in these extremes, but for a true normal distribution it has to be physically possible. For ages, a normally distributed model will predict there is a non-zero probability of data lying 5 standard deviations above or below the mean - which would correspond to physically impossible ages, such as below 0 or above 150. (Though if you look at a population pyramid, it's not clear why you would expect age to be even approximately normally distributed in the first place.) Similarly if you had heights data, which intuitively might follow a more "normal-like" distribution, it could only be truly normal if there were some chance of heights below 0 cm or above 300 cm.

I've occasionally seen it suggested that we can evade this problem by centering the data to have mean zero. That way both positive and negative "centered ages" are possible. But although this makes both negative values physically plausible and interpretable (negative centered values correspond to actual values lying below the mean), it doesn't get around the issue that the normal model will produce physically impossible predictions with non-zero probability, once you decode the modelled "centered age" back to an "actual age".

...so why bother testing? Even if not exact, normality can still be a useful model

The important question isn't really whether the data are exactly normal - we know a priori that can't be the case, in most situations, even without running a hypothesis test - but whether the approximation is sufficiently close for your needs. See the question is normality testing essentially useless? The normal distribution is a convenient approximation for many purposes. It is seldom "correct" - but it generally doesn't have to be exactly correct to be useful. I'd expect the normal distribution to usually be a reasonable model for people's heights, but it would require a more unusual context for the normal distribution to make sense as a model of people's ages.

If you really do feel the need to perform a normality test, then Kolmogorov-Smirnov probably isn't the best option: as noted in the comments, more powerful tests are available. Shapiro-Wilk has good power against a range of possible alternatives, and has the advantage that you don't need to know the true mean and variance beforehand. But beware that in small samples, potentially quite large deviations from normality may still go undetected, while in large samples, even very small (and for practical purposes, irrelevant) deviations from normality are likely to show up as "highly significant" (low p-value).

"Bell-shaped" isn't necessarily normal

It seems you have been told to think of "bell-shaped" data - symmetric data that peaks in the middle and which has lower probability in the tails - as "normal". But the normal distribution requires a specific shape to its peak and tails. There are other distributions with a similar shape on first glance, which you may also have characterised as "bell-shaped", but which aren't normal. Unless you've got a lot of data, you're unlikely to be able to distinguish that "it looks like this off-the-shelf distribution but not like the others". And if you do have a lot of data, you'll likely find it doesn't look quite like any "off-the-shelf" distribution at all! But in that case for many purposes you'd be just as well to use the empirical CDF.

Gallery of "bell shaped" distributions

The normal distribution is the "bell shape" you are used to; the Cauchy has a sharper peak and "heavier" (i.e. containing more probability) tails; the t distribution with 5 degrees of freedom comes somewhere in between (the normal is t with infinite df and the Cauchy is t with 1 df, so that makes sense); the Laplace or double exponential distribution has pdf formed from two rescaled exponential distributions back-to-back, resulting in a sharper peak than the normal distribution; the Beta distribution is quite different - it doesn't have tails that head off to infinity for instance, instead having sharp cut-offs - but it can still have the "hump" shape in the middle. Actually by playing around with the parameters, you can also obtain a sort of "skewed hump", or even a "U" shape - the gallery on the linked Wikipedia page is quite instructive about that distribution's flexibility. Finally, the triangular distribution is another simple distribution on a finite support, often used in risk modelling.

It's likely that none of these distributions exactly describe your data, and very many other distributions with similar shapes exist, but I wanted to address the misconception that "humped in the middle and roughly symmetric means normal". Since there are physical limits on age data, if your age data is "humped" in the middle then it's still possible a distribution with finite support like the Beta or even triangular distribution may prove a better model than one with infinite tails like the normal. Note that even if your data really were normally distributed, your histogram is still unlikely to resemble the classic "bell" unless your sample size is fairly large. Even a sample from a distribution like the Laplace, whose pdf is clearly distinguishable from that of the normal due to its cusp, may produce a histogram that visually appears about as similar to a bell as a genuinely normal sample would.

Normal and Laplace samples of various sample sizes

R code

plot(dnorm, -3, 3, ylab="probability density", main="Normal(0,1)") 
plot(function(x){dt(x, df=1)}, -3, 3, ylab="probability density", main="Cauchy") 
plot(function(x){dt(x, df=5)}, -3, 3, ylab="probability density", main="t with 5 df") 
plot(function(x){0.5*exp(-abs(x))}, -3, 3, ylab="probability density", main="Laplace(0,1)") 
plot(function(x){dbeta(x, shape1=2, shape2=2)}, ylab="probability density", main="Beta(2,2)")
plot(function(x){1-0.5*abs(x)}, -1, 1, ylab="probability density", main="Triangular")

normalhist <- function(n) {hist(rnorm(n), main=paste("Normal sample, n =",n), xlab="x")}
laplacehist <- function(n) {hist(rexp(n)*(1 - 2*rbinom(n, 1, 0.5)), main=paste("Laplace sample, n =",n), xlab="x")}

# No random seed is set
# Re-run the code to see the variability in histograms you might expect from sample to sample
normalhist(50); laplacehist(50)
normalhist(100); laplacehist(100)
normalhist(200); laplacehist(200)

Age can not be from normal distribution. Think logically: you cannot have negative age, yet normal distribution allows for negative numbers.

There are many bell-shaped distributions out there. If something looks bell-shaped it doesn't mean that it has to be normal.

There is no way to know for sure anything in statistics, including from which distribution the data comes from. The shape is a clue: bell-shape is one argument for normal distribution. Also, understanding your data is very important. The variable such as age is often skewed, which would rule out normality. As mentioned the normal distribution has no bounds, but it is sometimes used for bounded variables. For instance, if the mean age is 20 years, and the standard deviation is 1, then the probability of age <17 or >23 is less than 0.3%. So, it is possible than normal distribution could be a good approximation.

You can try running a statistical test for normality such as Jarque-Bera, which takes into account skewness and kurtosis of the sample. Kurtosis can be important in some cases. It's very important in finance, because if you model the data with normal distribution, but the data is in fact from a fat-tailed distribution, you may end up underestimating the risks and prices of assets.

It would help of you reported some descriptive statistics or a histogram of your age and height data, such as mean, variance, skewness, kurtosis.

  • $\begingroup$ Thank you for your help, can you tell me how to know that the certain data comes from normal distribution for example in your answer stated that age can not be from normal distribution, what about other data like height.What is the criteria that i must know.i want to learn more about this because it seems that i have misunderstand the concept since i am new to this.Thanks again. $\endgroup$
    – NoraNorad
    Commented Dec 17, 2014 at 6:43
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    $\begingroup$ Yet, normal distribution often is used as an approximation for such variables as age. And it is not really a problem since you can define age_centred as age - mean(age) and you have a variable with mean 0, with some standard deviation, positive and negative values. So I wouldn't be so strict about it. $\endgroup$
    – Tim
    Commented Dec 17, 2014 at 7:44
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    $\begingroup$ You cannot have negative height for people either, but that wouldn't be a barrier to me to describing height as normally distributed if that was a good approximation. For that matter, why use any distribution with infinite bounds for measurements that can only be finite? As @Tim says, it is all a matter of approximations acceptable given the data and given the purpose. $\endgroup$
    – Nick Cox
    Commented Dec 17, 2014 at 11:03
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    $\begingroup$ I agree that normal distribution could be a good approximation for bounded data sometimes, but the question was about whether the data is from normal or not. $\endgroup$
    – Aksakal
    Commented Dec 17, 2014 at 13:35
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    $\begingroup$ The age of graduating seniors from high school could potentially be normally distributed and also take on negative values if mean centered as @Tim mentioned. $\endgroup$
    – ui_90jax
    Commented Dec 19, 2015 at 18:55

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