The problem shows a uniform distribution of a random variable $X$ between 0 and 2. A uniform distribution has a constant probability density function (PDF) over its range.

For a uniform distribution over the interval $[a, b]$, the probability density function is: $f(x) = \frac{1}{b - a}$ where $a = 0$ and $b = 2$, so: $f(x) = \frac{1}{2}$

The probability $P(X > 0.72)$ can be calculated by determining the fraction of the interval from $0.72$ to 2. Since $X$ is uniformly distributed, the probability is proportional to the length of the interval.

The length of the interval from $0.72$ to 2 is: $2 - 0.72 = 1.28$

The total length of the distribution's interval is $2 - 0 = 2$. Therefore, the probability is: $P(X > 0.72) = \frac{1.28}{2} = 0.64$

Thus, the probability that $X > 0.72$ is $0.64$.

Do you have any further questions or need additional details?

Here are some related questions for further exploration:

1. What is the cumulative distribution function (CDF) of a uniform distribution?
2. How would you calculate $P(X < 1.5)$ for this distribution?
3. What is the expected value $E(X)$ for this uniform distribution?
4. How do you find the variance of a uniform distribution over the interval [0, 2]?
5. How does the probability change if the range of the uniform distribution was $[0, 4]$?

Tip: In a uniform distribution, probabilities are proportional to the length of the intervals within the distribution's support.