17
$\begingroup$

I've been in a debate with my graduate-level statistics professor about "normal distributions". I contend that to truly get a normal distribution one must have mean=median=mode, all the data must be contained under the bell curve, and perfectly symmetrical around the mean. Therefore, technically, there are virtually NO normal distributions in real studies, and we should call them something else, perhaps "near-normal".

She says I'm too picky, and if the skew/kurtosis are less than 1.0 it is a normal distribution and took off points on an exam. The dataset is total number of falls/year in a random sampling of 52 nursing homes which is a random sample of a larger population. Any insight?

Problem:

QUESTION: 3. Compute measures of skewness and kurtosis for this data. Include a histogram with a normal curve. Discuss your findings. Is the data normally distributed?

Statistics 
Number of falls  
N  Valid    52
   Missing   0
Mean        11.23
Median      11.50
Mode         4a

a. Multiple modes exist. The smallest value is shown

Number of falls  
N  Valid    52
   Missing   0
Skewness      .114
Std. Error of Skewness    .330
Kurtosis  -.961
Std. Error of Kurtosis    .650

My answer:

The data is platykurtic and has only slight positive skewing, and it is NOT a normal distribution because the mean and median and mode are not equal and the data is not evenly distributed around the mean. In reality virtually no data is ever a perfect normal distribution, although we can discuss “approximately normal distributions” such as height, weight, temperature, or length of adult ring finger in large population groups.

Professor's answer:

You are correct that there is no perfectly normal distribution. But, we are not looking for perfection. We need to look at data in addition to the histogram and the measures of central tendency. What do the skewness and kurtosis statistics tell you about the distribution? Because they are both between the critical values of -1 and +1, this data is considered to be normally distributed.

$\endgroup$
  • 3
    $\begingroup$ I would like to know your professor's exact wording. In principle a normal distribution has mean, median and mode identical (but so do many other distributions) and has skewness 0 and (so-called excess) kurtosis 0 (and so do some other distributions). At best a distribution with (e.g.) slight skewness or kurtosis is approximately normal. Note that almost all real data are at best approximations to named distributions in the theoretical menagerie. $\endgroup$ – Nick Cox Sep 17 '18 at 11:15
  • 22
    $\begingroup$ I don't agree with @user2974951 In company with every good text I know I am perfectly happy thinking that the normal distribution has a median and mode. And that applies widely to continuous distributions, although I don't doubt that pathological counter-examples can be identified. $\endgroup$ – Nick Cox Sep 17 '18 at 11:17
  • 4
    $\begingroup$ Thanks for the specific detail, which shows merit on both sides, but I am not grading either of you. However, I dissent strongly from the term critical values as used by your Professor, as the limits $\pm 1$ for skewness and kurtosis have no standing whatsoever beyond being rules of thumb that might be used. Depending on what you are doing with the data, a skewness $< 1$ might go along with wanting to transform the data and a skewness of $>1$ might go along with the leaving the data are they are, and similarly for kurtosis. $\endgroup$ – Nick Cox Sep 17 '18 at 12:39
  • 6
    $\begingroup$ If we seriously let ourselves embrace the art of nitpicking, we ought to observe that there are no negative falls, and that falls are discrete, so the distribution de facto cannot be normal. This renders the question void in the first place. On a more serious note, the question is clearly a fabricated example aiming to check specific rules of thumb. In reality, depending on the objective of our study, it may be reasonable or not to assume a normal distribution. In truth we will never know, as we only have a sample. $\endgroup$ – Ioannis Sep 17 '18 at 19:05
  • 5
    $\begingroup$ @user2974951 You ought to consider deleting your first comment, then, since you disagree with it now. So far it has fooled three readers into signaling they agree with it! $\endgroup$ – whuber Sep 17 '18 at 22:00
25
$\begingroup$

A problem with your discussion with the professor is one of terminology, there's a misunderstanding that is getting in the way of conveying a potentially useful idea. In different places, you both make errors.

So the first thing to address: it's important to be pretty clear about what a distribution is.

A normal distribution is a specific mathematical object, which you could consider as a model for an infinite population of values. (No finite population can actually have a continuous distribution.)

Loosely, what this distribution does (once you specify the parameters) is define (via an algebraic expression) the proportion of the population values that lies within any given interval on the real line. Slightly less loosely, it defines the probability that a single value from that population will lie in any given interval.

An observed sample doesn't really have a normal distribution; a sample might (potentially) be drawn from a normal distribution, if one were to exist. If you look at the empirical cdf of the sample, it's discrete. If you bin it (as in a histogram) the sample has a "frequency distribution", but those aren't normal distributions. The distribution can tell us some things (in a probabilistic sense) about a random sample from the population, and a sample may also tell us some things about the population.

A reasonable interpretation of a phrase like "normally distributed sample"* is "a random sample from a normally distributed population".

*(I generally try to avoid saying it myself, for reasons that are hopefully made clear enough here; usually I manage to confine myself to the second kind of expression.)

Having defined terms (if still a little loosely), let us now look at the question in detail. I'll be addressing specific pieces of the question.

normal distribution one must have mean=median=mode

This is certainly a condition on the normal probability distribution, though not a requirement on a sample drawn from a normal distribution; samples may be asymmetric, may have mean differ from median and so on. [We can, however, get an idea how far apart we might reasonably expect them to be if the sample really came from a normal population.]

all the data must be contained under the bell curve

I am not sure what "contained under" means in this sense.

and perfectly symmetrical around the mean.

No; you're talking about the data here, and a sample from a (definitely symmetrical) normal population would not itself be perfectly symmetric.

Therefore, technically, there are virtually NO normal distributions in real studies,

I agree with your conclusion but the reasoning is not correct; it's not a consequence of the fact that data are not perfectly symmetric (etc); it's the fact that populations are themselves not perfectly normal.

if the skew/kurtosis are less than 1.0 it is a normal distribution

If she said this in just that way, she's definitely wrong.

A sample skewness may be much closer to 0 than that (taking "less than" to mean in absolute magnitude not actual value), and the sample excess kurtosis may also be much closer to 0 than that (they might even, whether by chance or construction, potentially be almost exactly zero), and yet the distribution from which the sample was drawn can easily be distinctly non-normal.

We can go further -- even if we were to magically know the population skewness and kurtosis were exactly that of a normal, it still wouldn't of itself tell us the population was normal, nor even something close to normal.

The dataset is total number of falls/year in a random sampling of 52 nursing homes which is a random sample of a larger population.

The population distribution of counts are never normal. Counts are discrete and non-negative, normal distributions are continuous and over the entire real line.

But we're really focused on the wrong issue here. Probability models are just that, models. Let us not confuse our models with the real thing.

The issue isn't "are the data themselves normal?" (they can't be), nor even "is the population from which the data were drawn normal?" (this is almost never going to be the case).

A more useful question to discuss is "how badly would my inference be impacted if I treated the population as normally distributed?"

It's also a much harder question to answer well, and may require considerably more work than glancing at a few simple diagnostics.

The sample statistics you showed are not particularly inconsistent with normality (you could see statistics like that or "worse" not terribly rarely if you had random samples of that size from normal populations), but that doesn't of itself mean that the actual population from which the sample was drawn is automatically "close enough" to normal for some particular purpose. It would be important to consider the purpose (what questions you're answering), and the robustness of the methods employed for it, and even then we may still not be sure that it's "good enough"; sometimes it may be better to simply not assume what we don't have good reason to assume a priori (e.g. on the basis of experience with similar data sets).

it is NOT a normal distribution

Data - even data drawn from a normal population - never have exactly the properties of the population; from those numbers alone you don't have a good basis to conclude that the population is not normal here.

On the other hand neither do we have any reasonably solid basis to say that it's "sufficiently close" to normal - we haven't even considered the purpose of assuming normality, so we don't know what distributional features it might be sensitive to.

For example, if I had two samples for a measurement that was bounded, that I knew would not be heavily discrete (not mostly only taking a few distinct values) and reasonably near to symmetric, I might be relatively happy to use a two-sample t-test at some not-so-small sample size; it's moderately robust to mild deviations from the assumptions (somewhat level-robust, not so power-robust). But I would be considerably more cautious about as causally assuming normality when testing equality of spread, for example, because the best test under that assumption is quite sensitive to the assumption.

Because they are both between the critical values of -1 and +1, this data is considered to be normally distributed."

If that's really the criterion by which one decides to use a normal distributional model, then it will sometimes lead you into quite poor analyses.

The values of those statistics do give us some clues about the population from which the sample was drawn, but that's not at all the same thing as suggesting that their values are in any way a 'safe guide' to choosing an analysis.


Now to address the underlying issue with even a better phrased version of such a question as the one you had:

The whole process of looking at a sample to choose a model is fraught with problems -- doing so alters the properties of any subsequent choices of analysis based on what you saw! e.g for a hypothesis test, your significance levels, p-values and power are all not what you would choose/calculate them to be, because those calculations are predicated on the analysis not being based on the data.

See, for example Gelman and Loken (2014), "The Statistical Crisis in Science," American Scientist, Volume 102, Number 6, p 460 (DOI: 10.1511/2014.111.460) which discusses issues with such data-dependent analysis.

$\endgroup$
  • $\begingroup$ Hi Peter, sorry I didn't even see your post there. $\endgroup$ – Glen_b Sep 18 '18 at 9:36
  • $\begingroup$ This conversation has been moved to chat. $\endgroup$ – Glen_b Sep 19 '18 at 2:21
41
$\begingroup$

You're missing the point and probably are also being "difficult," which is not appreciated in the industry. She's showing you a toy example, to train you in assessment of normality of a data set, which is to say whether the data set comes from a normal distribution. Looking at distribution moments is one way to check the normality, e.g. Jarque Bera test is based on such an assessment.

Yes, the normal distribution is perfectly symmetrical. However, if you draw a sample from a true normal distribution, that sample will most likely not be perfectly symmetrical. This is the point you're completely missing. You can test this very easily yourself. Just generate a sample from Gaussian distribution, and check its moment. They'll never be perfectly "normal," despite the true distribution being such.

Here's a silly Python example. I'm generating 100 samples of 100 random numbers, then obtaining their means and medians. I print the first sample to show that the mean and median are different, then show the histogram of the difference between the means and medians. You can see that it's rather narrow, but the difference is basically never zero. Note, that the numbers are truly coming from a normal distribution.

code:

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(1)
s = np.random.normal(0, 1, (100,100))
print('sample 0 mean:',np.mean(s[:,0]),'median:',np.median(s[:,0]))

plt.hist(np.mean(s,0)-np.median(s,0))
plt.show()
print('avg mean-median:',np.mean(np.mean(s,0)-np.median(s,0)))

outputs: enter image description here

P.S.

Now, whether the example from your question should be considered normal or not depends on the context. In the context of what was taught in your class room you're wrong, because your professor wanted to see whether you know the rule of thumb test that she gave you, which is that skew and excess kurtosis need to be in -1 to 1 range.

I personally never used this particular rule of thumb (I can't call it a test), and didn't even know it existed. Apparently, some people in some fields do use it though. If you were to plug your data set descriptives into JB test, it would have rejected normality. Hence, you're not wrong to suggest that the data set is not normal, of course, but you're wrong in a sense that you failed to apply the rule that was expected from you based on what's been taught in the class.

If I were you I'd politely approach your professor and explain myself, as well as show JB test output. I'd acknowledge that based on her test my answer was wrong, of course. If you attempt to argue with her the way you argue here, your chances are very low to get the point back in the test, because your reasoning is weak about medians and means and samples, it shows lack of understanding of samples vs. populations. If you change your tune, then you'll have a case.

$\endgroup$
  • 23
    $\begingroup$ (+1) Exactly the point. Confusing the random variable and a sample of realisations from that random variable. $\endgroup$ – Xi'an Sep 17 '18 at 15:33
  • 15
    $\begingroup$ (+1) I would add that even if the population were not exactly normal, it typically doesn't matter. The real question is usually whether the data is "normal enough" that using certain tools/parameterizations makes sense. If the clouds opened up and a booming voice from the heavens says "THIS DATA IS DRAWN FROM A POPULATION WITH KURTOSIS 2.98", I wouldn't necessarily take that as a strong argument against using a $t$-test. $\endgroup$ – Matt Krause Sep 17 '18 at 17:39
  • $\begingroup$ Would it be fair to say that if your samples were perfectly normally distributed, that is string evidence that the samples aren't random? $\endgroup$ – JimmyJames Sep 18 '18 at 20:10
  • $\begingroup$ @JimmyJames, 4 years ago there was a paper in Science that claimed 20 minute conversation with a gay canvasser can change your feelings towards gay people. It turns out that the authors made up their survey data. They were too lazy and generated perfectly Gaussian noise, and that's how they were caught - see Irregularities in LaCour (2014) by Broockman et al $\endgroup$ – Aksakal Sep 18 '18 at 20:18
  • $\begingroup$ @Aksakal I'm not sure that's exactly the same thing as what i am asking. In that case I think the argument in that case was that real data is never perfectly normal. I'm starting from your statement "However, if you draw a sample from a true normal distribution, that sample will most likely not be perfectly symmetrical." If I'm randomly sampling from a perfect normal distribution, I would not expect that each successive data point will fall right in place where it needs to be to fill in a perfect normal curve. That would seem a lot like a non-random selection process to me. $\endgroup$ – JimmyJames Sep 18 '18 at 20:44
6
$\begingroup$

The teacher is clearly out of his/her element, and probably should not be teaching statistics. It seems worse to me to teach something wrong than to not teach it at all.

These issues could all be cleared up easily if the distinction between "data" and "process that produced the data" were made more clearly. Data target the process that produced the data. The normal distribution is a model for this process.

It makes no sense to talk about whether the data are normally distributed. For one reason, the data are always discrete. For another reason, the normal distribution describes an infinity of potentially observable quantities, not a finite set of specific observed quantities.

Further, the answer to the question "is the process that produced the data a normally distributed process" is also always "no," regardless of the data. Two simple reasons: (i) any measurements we take are necessarily discrete, being rounded off to some level. (ii) perfect symmetry, like a perfect circle, does not exist in observable nature. There are always imperfections.

At best, the answer to the question "what do these data tell you about normality of the data-generating process" could be given as follows: "these data are consistent with what we would expect to see, had the data truly come from a normally distributed process." That answer correctly does not conclude that the distribution is normal.

These issues are very easily understood by using simulation. Just simulate data from a normal distribution and compare those to the existing data. If the data are counts (0,1,2,3,...), then obviously the normal model is wrong because it does not produce numbers like 0,1,2,3,...; instead, it produces numbers with decimals that go on forever (or at least as far as the computer will allow.) Such simulation should be the first thing you do when learning about the normality question. Then you can more correctly interpret the graphs and summary statistics.

$\endgroup$
  • 10
    $\begingroup$ I didnt downgrade your answer, but consider that you're judging a grad professor from the words of a student. How likely it is that a student is right and a teacher is wrong? Isn't it more likely that student is misrepresenting his professor and the context of the conversation? $\endgroup$ – Aksakal Sep 17 '18 at 15:19
  • $\begingroup$ Based on my experience, and on the students words, I'd say it is more likely that the teacher is wrong. There are teachers with little formal training that teach courses, even grad courses, in universities everywhere. If the accreditating agencies only knew the truth! $\endgroup$ – Peter Westfall Sep 17 '18 at 15:54
  • 6
    $\begingroup$ @Possum-Pie, I can guess what is expected from you. It's probably 101-ish course in stats, so you have to look at skewness and kurtosis. If they're not close enough to 0 and 3, then you say it's not normal. That's all. In fact that's what JB test does in a more formal way. The point of the exercise is for you to remember that Gaussian has skew 0 and kurtosis 3. You're turning this silly but necessary exercise into a philosophical discussion. $\endgroup$ – Aksakal Sep 17 '18 at 16:56
  • 2
    $\begingroup$ The teacher's comment "Because they are both between the critical values of -1 and +1, this data is considered to be normally distributed" definitely either shows (i) lack of understanding or (ii) willingness to teach that which s/he knows to be wrong. I don't think that it is a philosophical discussion to question teacher preparedness or pedagogical methods. $\endgroup$ – Peter Westfall Sep 17 '18 at 20:24
  • 3
    $\begingroup$ "Consistency" language is good. But as Possum-Pie noted, teachers tell students, "based on this test/diagnostic, the data are normal," which is wrong on several counts. Teachers (psych and otherwise) need to (i) distinguish data-generating process from data, (ii) tell students that the normal and other models are models for the data-generating process, (iii) tell them that the normal distribution is always wrong as a model, regardless of the diagnostics, and (iv) tell them that the point of the exercise is to diagnose degree of non-normality, not answer yes/no. Then explain why it matters. $\endgroup$ – Peter Westfall Sep 17 '18 at 21:35
4
$\begingroup$

I'm an engineer, so in my world, the applied statistician is what I see most, and get the most concrete value. If you are going to work in applied, then you need to be solidly grounded in practice over theory: whether or not it is elegant, the aircraft has to fly and not crash.

When I think about this question the way I approach it, as many of my technical betters here have also done, is to think about "what does it look like in the real world with the presence of noise".

The second thing that I do is, often, to make a simulation that allows me to get my hands around the question.

Here is a very brief exploration:

#show how the mean and the median  differ with respect to sample size

#libraries
library(reshape2)
library(ggplot2)

#sample sizes
ssizes <- 10^(seq(from=1, to=3, by=0.25))
ssizes <- round(ssizes)

#loops per sample
n_loops <- 5000

#pre-declare, prep for loop
my_store <- matrix(0, 
                   ncol = 3, 
                   nrow = n_loops*length(ssizes))

count <- 1

for(i in 1:length(ssizes)){

  #how many samples
  n_samp <- ssizes[i]

  for(j in 1:n_loops){

    #draw samples
    y <- 0
    y <- rnorm(n = n_samp,mean = 0, sd = 1)

    #compute mean, median, mode
    my_store[count,1] <- n_samp
    my_store[count,2] <- median(y)
    my_store[count,3] <- mean(y)


    #update
    count = count + 1
  }
}


#make data into ggplot friendly form
df <- data.frame(my_store)
names(df) <- c("n_samp", "median","mean")

df <- melt(df, id.vars = 1, measure.vars = c("median","mean"))


#make ggplot
ggplot(df, aes(x=as.factor(n_samp), 
               y = value, 
               fill = variable)) + geom_boxplot() + 
  labs(title = "Contrast Median and Mean estimate variation vs. Sample Size",
       x = "Number of Samples",
       y = "Estimated value")

It gives this as the output: enter image description here

Note: be careful about the x-axis, because it is log-scaled, not uniform-scaled.

I know that the mean and median are exactly the same. The code says it. The empirical realization is greatly sensitive to sample size, and if there aren't truly infinite samples, then they can't ever perfectly match with theory.

You can think about whether the uncertainty in the median envelopes the estimated mean or vice versa. If the best estimate of the mean is within the 95% CI of the estimate for the median, then the data can't tell the difference. The data says they are the same in theory. If you get more data, then see what it says.

$\endgroup$
  • 1
    $\begingroup$ Interesting graph. I would have thought the Mean would have been generally larger than the median at first considering the mean chases outliers...in other words the red bars would be mean and green be medians. What am I missing? $\endgroup$ – Possum-Pie Sep 17 '18 at 13:55
  • 1
    $\begingroup$ @Possum-Pie Remember that outliers can be in either direction... the normal distribution has both a left tail and a right tail! $\endgroup$ – Silverfish Sep 17 '18 at 20:58
  • 2
    $\begingroup$ @Will that's a pretty standard implementation of a boxplot. $\endgroup$ – Glen_b Sep 19 '18 at 12:43
  • 1
    $\begingroup$ @Glen_b I have seen plenty of textbooks that do not teach the use of dots for outliers, so can understand someone not being used to them. But according to Hadley, the dots have been there even when Tukey introduced his "schematic plot" in 1970. $\endgroup$ – Silverfish Sep 20 '18 at 21:19
  • 1
    $\begingroup$ Yeah, a version without outliers in it (just based on a 5 number summary) would essentially be Mary Spear's range plot (1952). (NB that paper misses some important historical boxplot precursors, pre 1952) $\endgroup$ – Glen_b Sep 21 '18 at 0:27
4
$\begingroup$

In medical statistics, we only ever comment on the shapes and seeming of distributions. The fact that no discrete finite sample can ever be normal is irrelevant and pedantic. I would mark you wrong for that.

If a distribution looks "mostly" normal, we are comfortable with calling it normal. When I describe distributions for a non-statistical audience, I am very comfortable with calling something approximately normal even when I know the normal distribution is not the underlying probability model, I get the sense I would side with your teacher here... but we have no histogram or dataset to verify.

As a tip, I would go through the following inspections very closely:

  • who are the outliers, how many and what are their values?
  • Are the data bimodal?
  • Do the data seem to take a skewed shape so that some transformation (like a log) would better quantify the "distance" between observations?
  • Is there apparent truncation or heaping so that assays or labs are failing to reliably detect a certain range of values?
$\endgroup$
  • $\begingroup$ It seems in a field with so much math, people would be more strict between saying something is "normal distribution" which has certain very strict conotations, and saying it is "nearly normal". I'd never say that 1.932 is 2. but I may say it is nearly 2. $\endgroup$ – Possum-Pie Sep 17 '18 at 17:00
  • 1
    $\begingroup$ "Irrelevant and pedantic"? Seriously? I agree with Possum-Pie. I also would never say that 1.932 is the same as 2.0. Saying that data are "normal" confuses everything, from the meaning of the normal distribution as a model for the process that produced the data, to the real fact that normal distributions never precisely model our processes. Everyone should be taught that when they learn the normal distribution so they don't make silly statements. $\endgroup$ – Peter Westfall Sep 17 '18 at 20:30
  • 2
    $\begingroup$ @PeterWestfall I think part of the issue here is that "the data come from a normal distribution" is almost never literally true, and even if it were true, it would likely be impossible to prove it conclusively. So since the phrase would hardly ever be literally true, people will instead use "the data is normal" as a convenient short-hand to mean "the data seem close enough to normality for practical purposes" or "the normal distribution is a good-enough model for our DGP". $\endgroup$ – Silverfish Sep 17 '18 at 20:51
  • $\begingroup$ So why teach what is wrong when it is so simple to teach what is right? $\endgroup$ – Peter Westfall Sep 17 '18 at 20:53
  • 3
    $\begingroup$ @PeterW The linguistic point isn't just about the teaching, it's about the way the phrase is used (and intended to be construed) in everyday life: "the data is normal" is almost never used to mean "I know for certain that the population the data was sampled from is normal", because it could hardly ever meant that. It would be nicer if people said "the data seems normal" or even "the data looks normalish" (i.e. seems close enough to normal that we don't care about its deviation from normality) but particularly in an applied setting people will often say things like that. $\endgroup$ – Silverfish Sep 17 '18 at 21:47
2
$\begingroup$

I think you and your professor are talking in different context. Equality of mean = median = mode is characteristics of theoretical distribution and this is not the only characteristics. You can not say that if for any distribution above property hold then distribution is normal. T-distribution is also symmetric but it is not normal. So, you are talking about theoretical properties of normal distribution which hold always true for normal distribution.

You professor is talking about distribution of sample data. He is right, you will never get data in real life, where you will find mean = median = mode. This is simply due to sampling error. Similarly, it is very unlikely, you will get zero coefficient of skewness for sample data and zero excess kurtosis. Your professor is just giving you simple rule to get an idea about the distribution from the sample statistics. Which is not true in general (without getting further information).

$\endgroup$
  • 3
    $\begingroup$ Professor is said to be femaie. $\endgroup$ – Nick Cox Sep 17 '18 at 11:20
  • $\begingroup$ Why you don't get mean=median=mode is mostly because many distributions really are skewed! (Strictly, mean=median=mode is possible with skewed distributions too, despite what many textbooks say.) $\endgroup$ – Nick Cox Sep 17 '18 at 11:53
  • 1
    $\begingroup$ I disagree that lack of equality of mean/median/mode= sampling error. Suppose you random sampled 52 nursing homes for fall rates. Homes 27, 34, and 52 are chronicly short-staffed and always have above-average number of falls. Those homes pushes mean towards tail and isn't due to a sampling error. $\endgroup$ – Possum-Pie Sep 17 '18 at 11:54
  • 1
    $\begingroup$ @Possum Pie What the data are is secondary here but you're giving different signals in different places. Here you are talking about several nursing homes -- but in your question you state "in a nursing home". Being unclear about even incidental details does not help. $\endgroup$ – Nick Cox Sep 17 '18 at 11:59
  • $\begingroup$ @Nick Cox Sorry, I clarified it. Number of falls/year in a sample of 52 nursing homes $\endgroup$ – Possum-Pie Sep 17 '18 at 12:12
1
$\begingroup$

For practical purposes, underlying processes such as this one are usually finely approximated by normal distribution without anyone raising an eyebrow.

However, if you wanted to be pedantic the underlying process in this case can't be normally distributed, because it can't produce negative values (number of falls can't be negative). I wouldn't be surprised if it was in fact at least a bi-modal distribution with second peak close to zero.

$\endgroup$
  • $\begingroup$ It IS bimodal with modes at 4 falls and 13 falls. There are no zero falls reported. $\endgroup$ – Possum-Pie Sep 17 '18 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.