Why don't we automatically have outliers when mean and median differ strongly? Assume you have a data set with information on income of all students in the lecture. The mean value is 1500\$. The median value is however only 800\$. Which of the following conclusions is wrong?
The answer is: The difference between median and mean can be explained by outliers.
Why is this wrong? When Mean and Median deviate in such a manner, there have to be outliers or am I completely wrong? I can check outliers with |z| > 2.5 or a boxplot with IQR*1.5
But why do there not have to be outliers in such a case? I am completely lost and cannot think of a data set that fulfils such a case. Could you please help me?
 A: Sometimes a sample with a mean larger than the sample median can have boxplot outliers, as you say. (Sometimes many outliers.) The exponential sample with $n=1000$ observations below illustrates this. [Using R.]
set.seed(128)
y = rexp(1000)
mean(y); median(y)
[1] 1.029919
[1] 0.679321
boxplot(y, horizontal=T)
 abline(v=mean(y), col="red")

In the figure below, the heavy black line within the box shows the
sample median $0.6793$ and the vertical red line shows the sample mean $1.0299.$

The population mean is $\mu = 1$ and the population median is $\eta = -\ln(.5) = 0.6931.$
qexp(.5); -log(.5)
[1] 0.6931472
[1] 0.6931472

Below is a histogram of the $1000$ observations along with
the density curve of the distribution $\mathsf{Exp}(1).$
hist(y, prob=T, col="skyblue2")
 curve(dexp(x, 1), add=T, lwd=2, col="orange")


By contrast, moderately large samples from many right-skewed distributions have
mean larger than median, but may not have any outliers at all.
Many examples can be found among samples from beta and
other distributions.
set.seed(2022)
x = rbeta(1000, 2, 4)
mean(x); median(x)
[1] 0.3369563      # sample mean
[1] 0.3059188      # sample median
boxplot(x, horizontal=T)
 abline(v=mean(x), col="red")


The population mean is $\mu = 1/3$ and the population median is $\eta = 0.3138.$
qbeta(.5, 2, 4)
[1] 0.3138102

hist(x, prob=T, col="skyblue2")
 curve(dbeta(x, 2, 4), add=T, lwd=2, col="orange")


A: Here is an example. Say there are 2001 students, and 1000 of them have a score of 0 and 1001 of them have a score of 100. Then the mean is around 50, but the median is 100, and there are no outliers in this dataset.
A: Outliers from what? You must be assuming a normal (Gaussian) distribution. But with lognormal distributions, some high values are expected as the distribution can be quite skewed. The mean is almost always larger than the median with data sampled from lognormal distributions.  To test this idea, take the logarithm of all the values, and see if that distribution is closer to normal (Gaussian).
A: The difference between the mean and the median is not a good sign of outliers.
Example: You have five students, four earning \$1 and the fifth earning \$6 dollars. Then, the median income is \$1 and the average income is \$2. In this example, the average is two times the median! But that doesn't imply we should exclude one student from our analysis.
