Example distribution where 74% of probability is above the mean Watching Why You Should Want Driverless Cars On Roads Now, at 8:14 Derek Muller claims:

Surveys show 74 % of people believe they are above average drivers.

This claim motivates my question, but some clarification is needed. I am not asking for someone to source out the statistic nor to address the empirical claim. Rather, I would like a concrete and simple example of my intuition that it is mathematically possible. I've put together a wish list of additional criteria:

*

*Has a 'clean' algebraic expression for its probability density function or probability mass function (i.e. not just a constructed dataset or a 'tortured piecewise function')

*Support is single-variable

*Support is continuous or contiguous

*If support is discrete, each bin with nonzero probability mass must have one or two adjacent bins with nonzero probability mass

*Distribution is unimodal

 A: This is possible if there are very large outliers.
In general, if a distribution has outliers with values that are extremely different from the values of the rest of the distribution, then you are likely to have a mean significantly different from the median value.
So, for instance, a distribution of [-1000, 1, 2, 3] would have a median value of 1.5, a mean value of -248.5, and 75 percent of its distribution would be above the mean value.
A: The probability of being below the mean for a Poisson distribution with $\lambda=1$ is $73.57$6%. So it shouldn't be hard to imagine a continuous distribution that gets that number over $74$% (you asked for above the mean, but that can be solved with a simple transformation). Pareto distributions can easily have more than $74$% of the probability mass on one side of the mean. For instance, $x_m=2, \alpha = 1$ gives $75$% below the mean.
A: Consider household income distribution in the United States. The mean for that distribution is about $72K, which is much larger than the median. You can estimate what fraction of the distribution is less than the mean from this image. If it's not as much as 75% you can skew it as much as you like by adding a few more high earners.

https://www.census.gov/library/visualizations/2015/demo/distribution-of-household-income--2014.html
A: A simple real world example would be the number of legs that people have. The huge majority have two legs, nobody has three legs, some few people have one leg or none. The average is just below 2, and the huge majority is a tiny bit above average.
The opposite: Number of citizenships that people have. The huge majority have one, a very small number have none, and a bit more have two or more. The average is just a bit over 1, and the huge majority is below average.
Profit from a risk-only life insurance for the insurance company: If you pay for a life insurance that pays on death only, there will be no payout for most people, with a small profit for the insurance company. A small number create a huge loss. So the majority will create much higher than average profit. This is the situation from another answer, with huge negative outliers, only with a real example.
A: The existing answer seems to meet all the requirements in your wish list.  For completeness, I just thought I'd add an extremely simple case matching the main requirement, without using continuity, etc.  This example does not meet your criteria, but it is added to show that it is extremely easy to get distributions that meet  the main requirement (which may be counter-intuitive to some readers).

Example: If $X \sim \text{Bern}(\theta)$ with non-degenerate parameter $0<\theta<1$ you get:
$$\mathbb{P}(X > \mathbb{E}(X)) = \mathbb{P}(X > \theta) = \mathbb{P}(X = 1) = \theta.$$
Taking $\theta = 0.74$ then gives you the required outcome $\mathbb{P}(X > \mathbb{E}(X)) = 0.74$.
A: A beta distribution should satisfy your criteria. With first shape parameter of 1.0 and second shape parameter of 0.14, the average is 0.8772 while 74.56 % of the probability is concentrated above that.
Here is a graph of the pdf: 
