To solve the problem using the provided density function, let's follow these steps:

1. Identify the relevant interval and height:

   • The density function is a uniform distribution over the interval [2, 7].
   • The height of the density function (which represents the probability density) is 0.2.

2. Calculate the probability $P(X < 4.52)$:

   • This is the area under the density function from $X = 2$ to $X = 4.52$.

3. Compute the area:

   • The base of the rectangle (interval length) is $4.52 - 2 = 2.52$.
   • The height of the rectangle is given as 0.2.
   • Therefore, the probability is given by: $P(X < 4.52) = \text{base} \times \text{height} = 2.52 \times 0.2 = 0.504$

4. Round the answer to two decimal places: $P(X < 4.52) \approx 0.50$

So, the probability $P(X < 4.52)$ is approximately 0.50.

Would you like more details or have any further questions? Here are some related questions you might find useful:

1. What is the probability $P(X > 5)$ given the same density function?
2. How would you calculate the probability $P(3 < X < 6)$ using this density function?
3. What is the cumulative distribution function (CDF) for this uniform distribution?
4. How would the probability change if the interval for the density function was [1, 8]?
5. What is the expected value (mean) of this distribution?

Tip: The area under the curve of a probability density function (PDF) for continuous distributions represents the probability for a given interval.