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In the context of a probability density function (PDF), the mean time is an essential concept. It signifies the average time taken for an event, such as the completion of an exam in this exercise.
The mean time, often denoted by \( \mu \), provides an expectation of the central tendency for continuous random variables described by a PDF. To calculate the mean time, we use the formula:

• \( \mu = \int_{a}^{b} x p(x) \, dx \)

This integral calculates the expected value by weighing the time \( x \) by the density function \( p(x) \) over the interval. For our specific problem where \( p(x) = \frac{x^3}{4} \), the mean time is computed over the interval from 0 to 2 hours. By integrating \( x \times \frac{x^3}{4} \) from 0 to 2, we find the antiderivative, \( \frac{x^5}{20} \), and evaluate it to discover that the mean time of exam completion is 1.6 hours.
This value indicates that, on average, students take 1.6 hours to finish the exam. Calculating the mean is useful for understanding the central inclination of the data.

The median represents the midpoint of a probability distribution, where half the data lies below it and half above. In a continuous distribution with a PDF, finding the median involves determining the value \( m \) such that the area under the curve from the beginning of the interval to \( m \) equates to 50% of the total area.
To calculate the median for a distribution defined by \( p(x) = \frac{x^3}{4} \), set up the equation:

• \( \int_{0}^{m} \frac{x^3}{4} \, dx = 0.5 \)

This integral aims to find \( m \) where the cumulative probability up to that point is 0.5. In our problem, the antiderivative \( \frac{x^4}{16} \) is used, and solving \( \frac{m^4}{16} = 0.5 \) gives \( m^4 = 8 \), resulting in \( m = \sqrt[4]{8} \approx 1.68 \).
The median, approximately 1.68 hours, tells us the point at which exactly half of the students complete their exams within this time. The median provides a robust measure, giving insights where mean might be skewed due to extreme values.

Integration is a fundamental mathematical concept used to calculate areas under curves. This is particularly useful in probability density functions where integration helps find probabilities over intervals.
A PDF defines how probabilities are distributed over an interval, and integrating this function over a certain range yields the total probability for that range. For example, in our exercise, to find the proportion of students finishing their exam between 1.5 and 2.0 hours, we integrate the function \( \frac{x^3}{4} \) between these limits:

• \( \int_{1.5}^{2.0} \frac{x^3}{4} \, dx \)

The antiderivative, \( \frac{x^4}{16} \), is evaluated from 1.5 to 2.0, which yields a result of approximately 0.6836. This means about 68.36% of students will finish their exam in this timeframe.
Understanding integration is crucial as it allows us to move from simple probability density functions to practical applications, aiding not only in probability but also in fields such as physics and engineering.