# Theoretical Median and Mean question [closed]

How likely is this scenario?

In a class of 100 people who take an exam, the median of the exam is 82 and the mean is 77. What if the sample size is cut in half and the median/mean stay the same?

So how likely is the initial scenario and how does it change when the sample size is reduced or increased.

Thanks for any help.

Edit: I would think that a difference between the median and mean for a sample of that size would be unlikely in the first place?

## closed as off-topic by Watson, Laurent Duval, Claude Leibovici, Behrouz Maleki, user223391 Dec 18 '16 at 18:22

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• Can you please update your question with your own thoughts? People here don't usually like to do your homework for you without any participation on your side... – gt6989b Dec 15 '16 at 16:04
• Just did, can you comment on my assertion? – NoobyHacker Dec 15 '16 at 16:18
• I don't know whether this is hwk or just exploring a 'what if' on your own. But it is a very nice and original (if somewhat vague) question, which I was happy to see. – BruceET Dec 15 '16 at 21:48

There is no use speculating about this, when one can try the experiment repeatedly, using simulation, and see what happens.

First, here are 100 'test scores' for which the mean 72.37 and the median is 78.

  33 38 40 41 41 43 43 46 48 48 50 50 50 50 52 52 52 53 53 54 55 55 56 56 59
69 70 70 71 72 72 73 73 73 74 75 75 76 76 77 77 77 78 78 78 78 78 78 78 78
78 78 79 79 80 80 80 80 80 80 80 80 80 80 81 81 81 81 81 81 81 81 81 82 82
83 83 83 83 83 84 84 84 84 84 85 85 85 85 86 86 86 86 86 87 88 88 90 90 90


I put this sample of size $n = 100$ into a vector x. Then I took 10,000 random sub-samples (without replacement), each of size 50. For each subsample, I found the mean a and the median h.

At the end, I asked what percentage of the subsamples have mean less than median. Surprisingly, that was true 100% of the time.

There is no guarantee that it must always happen, but it seems very likely that the mean will be less than the median when you try this in a real situation. [(a) The smallest median in any subsample was 72.5 and the largest mean in any subsample was 77.86. But every subsample of 50 had a smaller mean than median. (b) For a different, slightly less skewed x with mean 74.85 and median 78, I got one subsample out of 10,000 with mean larger than median.]

The code in R statistical software that I used for this simulation is shown below:

 m = 10^4;  a = h = numeric(m)
for (i in 1:m) {
y = sample(x, 50)
a[i] = mean(y);  h[i] = median(y) }
mean(a < h)  # proportion with mean < median
## 1


Here are histograms of the 10,000 differences 'Median - Mean'. Notice that all differences are positive. Note: Following @jjet's Comment, which mirrors my own teaching experience, here is how I got my original sample x of 100 'scores': I took a sample of size 75 from $Norm(\mu = 80, \sigma=5)$, for students paying some attention, and a sample of size 25 from $Norm(50, 10)$, for students not paying so much attention, and then merged them into a 'class' of 100.

• Do you think if you used an initial size of scores like 500 and then took sub samples of size (150) this would change? I guess I can test it out myself but just curious if there is any type of theory behind it – Brandon Dec 15 '16 at 17:50
• Actually, I think the distribution of the scores would be more telling than the sample size. I'm going to experiment – Brandon Dec 15 '16 at 18:27
• @Brandon: Here view the class as the pop. The sample mean $\bar X_n = A_n$ approaches the pop mean $\mu$ as $n$ increases, and the dist'n of $A_n$ is nearly normal for moderate $n$. So we can pretty much predict how close $A_n$ is to $\mu.$ Similarly, the samp median $\tilde X_n = H_n$ approaches the pop median $\eta$ as $n$ increases. Also here $H_n$ is nearly normal. (There is a CLT for medians also.). So if $\mu$ and $\eta$ are substantially different, values of $A_n$ and $H_n$ will cluster ever more tightly around different values as $n$ increases. – BruceET Dec 15 '16 at 21:36
• @Brandon: Changed notation slightly. Just noticed bar -  and tilde ~ look a lot the same in this small type. – BruceET 13 mins ago – BruceET Dec 15 '16 at 21:37

Suppose $X$ is a random variable which denotes the marks in the exam. Given a sample $\{X_1, X_2, \ldots , X_n\}$ of size $n$, the mean and median are both well known measures of central tendency of the distribution of $X$. Since it is random variables we're talking about, and the sample mean and sample median are actually realizations of a random process, we do not have a hold on the exact values of these quantities. We can only handle them in terms of chance, i.e. probability. As $n$ increases, we expect the mean and median to get closer. But in reality, they only get closer in terms of chance, i.e. for some fixed pre-specified positive real number $\epsilon > 0$,

$$P_n(\epsilon)=Pr(|\overline{X_n}-X_n^{med}|>\epsilon)$$

becomes smaller as $n$ increases (Of course the quantity will depend on the underlying distribution of $X$). So even if you consider a sample of size $10000$ instead of one of size $100$, it is still possible that the realized value of sample mean and sample median differs a lot. But the probability that they will differ more than some pre-specified quantity will be smaller. That said, there are some subtle features that really make these measures different. For example, median is a robust estimator, while mean is not. If you change a sample value and make it very large or very small, the mean will be affected badly, but the median remains same, and hence the median is said to be more stable than the mean. Hope that helps.

• So it is very possible that there Is such a large difference between the median and mode? – NoobyHacker Dec 15 '16 at 16:26
• Yes. It is ALWAYS possible. But loosely speaking, the chance (in terms of probability) decreases as $n$ increases. – user398842 Dec 15 '16 at 16:31
• @aditi_ray: For a sample from a normal distribution the sample mean and median do get closer as $n$ increases. But the point here is we doubt the population is normal, instead believing it is left-skewed so the pop mean is below the pop median. In that case, as $n$ increases, the sample mean approaches the pop mean, and the sample median approaches the pop median, so the separation btw sample mean and median becomes more distinct. – BruceET Dec 15 '16 at 21:08
• @BruceET: That's a valid point. When the population mean and population variance are distinct, then $P_n(\epsilon)$ defined above will tend to $0$ as $n$ goes up. However, if we do not have any prior information about the distribution, we can employ the powerful central limit theorem to re-scale the data and assume normality. – user398842 Dec 16 '16 at 4:06

I think this question is tougher than it seems at first glance. However, if you randomly sample a population, you would not change the theoretical mean or median. That can be proved with a little technical setup.

You mentioned though that this is a finite class with 100 students and not a theoretically infinite population. Therefore, the sample mean and median are estimators which would likely have non-neglible variances. If the sample mean and median for the whole class is 82 and 77, then recalculating these estimators on half the class would almost certainly alter their values.

*I'm new here so I can't comment directly on BruceET's answer. But I'll add that the simulation he did is highly dependent on the distribution of the underlying data. Thus, speculation is all that we can really do here. To emphasize this point, consider what would we redefined the original R data as

x = c(rep(44.5102, 49), rep(78, 2), rep(100, 49))

Then, the mean and median are the if we run the same simulation, the mean will only be less than the median 49% of the time -- not 100%.

• But hypothetically speaking, is that much of a difference between the median and the mean likely? – NoobyHacker Dec 15 '16 at 16:19
• It's extremely likely. I've taught countless stat classes and I always see a strongly left-skewed distribution of scores. That causes the mean to be considerably lower than the median. I just checked the last two midterms I graded and the mean was 6 points lower than the median both times. – jjet Dec 15 '16 at 16:26
• @jjet. I don't think we have a disagreement. My x was patterned after your idea of a left-skewed distribution where pop mean and pop median differ. Maybe a better simulation would have been to generate a different class x at each iteration according to something like 75% NORM(80,5) and 25% NORM(50,10) mixture. – BruceET Dec 15 '16 at 21:42