mathematical-statisticsmeanmedian
It is a known fact that median is resistant to outliers. If that is the case, when and why would we use the mean in the first place?
One thing I can think of perhaps is to understand the presence of outliers i.e. if the median is far from the mean, then the distribution is skewed and perhaps the data needs to be examined to decide what is to be done with the outliers. Are there any other uses?
You are asking an applied and subjective question, and thus, any answer needs to be infused with applied and subjective considerations.
From a purely statistical perspective, the mean and median both provide different information about the central tendency of a sample of data. Thus, neither is correct or incorrect by definition.
From an applied perspective, we often want to say something meaningful about the central tendency of a sample, where central tendency maps onto some subjective notion of "typical".
When summarising what is typical in a sample, observations that are many standard deviations away from the mean (perhaps 3 or 4 SD) will have a large influence on the mean, but not the median. Such observations may lead the mean to deviate from what we think of as the "typical" value of the sample. This helps to explain the popularity of the median when it comes to reporting house prices and income, where a single island in the pacific or billionaire could dramatically influence the mean, but not the median. Such distributions can often include extreme outliers, and the distribution is positively skewed. In contrast, the median is robust.
The median can be problematic when the data takes on a limited number of values. For example, the median of a 5-point Likert item lacks the nuance possessed by the mean. For example, means of 2.8, 3.0, and 3.3 might all have a median of 3.
In general, the mean has the benefit of using more of the information from the data.
When skewed distributions exist, it is also possible to transform the distribution and report the mean of the transformed distribution.
When a distribution includes outliers, it is possible to use a trimmed mean, or remove the outliers, or adjust the value of the outlier to a less extreme value (e.g., 2 SD from the mean).
A histogram represents probability by area:
In this figure, the white region (to the left of $x=1$) comprises half the area. The blue region comprises the other half. The boundary between them at $x=1$ is, by definition, the median: it splits the total probability exactly in half.
The areas in the next figure are shaded with varying densities of black:
The density of black is directly proportional to the horizontal distance from the middle (around 1.65 here). Each point near $x=7$ is very dark. Such points contribute proportionately more to the total amount of black ink used to shade this figure. The central place (where the shading becomes white) is chosen to make total amount of black to its right equal the total amount of black to its left. This makes it equal to the mean.
We see that the distant values ($x$ larger than $3$ or so) contribute so much black that they "pull" the dividing line--the mean--towards them.
Another way to see this uses three dimensions. The mean is the point at which the two volumes (pink/yellow and blue/purple) are exactly equal:
This figure was constructed by sweeping the original histogram (shown in the $x$ (horizontal) and $z$ (up) directions) from side to side around the mean value. This caused the long extended "tail" at the right to sweep out a larger region, because it is further from the mean than the rest of the figure. By virtue of that, it contributes more to the volume.
Were we to try the same thing by sweeping around the median (at $x=1$), we would get unequal volumes:
The white line on the ground still shows the mean, but now the axis of sweeping is around the median. Although the median correctly splits the cross-sectional area into two, it allows more volume to the right because the points to the right are "skewed" away from the median. Thus the sweeping axis has to be shifted toward larger values of $x$ to make the volumes balance.
Best Answer
In a sense, the mean is used because it is sensitive to the data. If the distribution happens to be symmetric and the tails are about like the normal distribution, the mean is a very efficient summary of central tendency. The median, while being robust and well-defined for any continuous distribution, is only $\frac{2}{\pi}$ as efficient as the mean if the data happened to come from a normal distribution. It is this relative inefficiency of the median that keeps us from using it even more than we do. The relative inefficiency translates into a minor absolute inefficiency as the sample size gets large, so for large $n$ we can be more guilt-free about using the median.
It is interesting to note that for a measure of variation (spread, dispersion), there is a very robust estimator that is 0.98 as efficient as the standard deviation, namely Gini's mean difference. This is the mean absolute difference between any two observations. [You have to multiply the sample standard deviation by a constant to estimate the same quantity estimated by Gini's mean difference.] An efficient measure of central tendency is the Hodges-Lehmann estimator, i.e., the median of all pairwise means. We would use it more if its interpretation were simpler.