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A Cumulative Distribution Function (CDF) tells us the probability that a random variable, say \(X\), is less than or equal to a certain value \(x\). It's a useful tool for understanding the probability spread over the range of possible outcomes.

For the given problem, the CDF is defined piecewise for the amount of time a book is checked out. Specifically, it's zero before the book is checked out (\(x < 0\)), increases as \(\frac{x^2}{4}\) while the book is checked out (\(0 \leq x < 2\)), and then it reaches one once the maximum checkout time happens (\(x \geq 2\)).

This setup allows us to calculate specific probabilities, such as the ones listed in the exercise, by substituting the values of \(x\) into the CDF. This can cover probabilities of the form \(P(X \leq a)\), \(P(a \leq X \leq b)\), and so on.

The Expected Value of a continuous random variable provides a measure of its central tendency, averaged over its range of values. It's represented as \(E(X)\).

For our problem, the expected duration of a book checkout, where the PDF is given as \(f(x) = \frac{x}{2}\) between 0 and 2, is found by integrating the product of \(x\) and \(f(x)\) across all possible values of \(x\).

This process mathematically translates into:

• \[ E(X) = \int_{0}^{2} x \cdot \frac{x}{2} \, dx \]
• Which results in \(\frac{1}{2} \int_{0}^{2} x^2 \, dx = \frac{4}{3}\).

This value (\(\frac{4}{3}\)) is the mean time for which a book is checked out, allowing students to understand on average how long the book is loaned.

Variance provides a measure of the spread of a set of values, indicating how much they deviate from the expected value. The standard deviation is simply the square root of the variance, offering a scale measurement of this spread.

In the context of our exercise, the variance \(V(X)\) is calculated by first determining \(E(X^2)\), which involves integrating the square of \(x\) against its PDF:

• \[ E(X^2) = \int_{0}^{2} x^2 \cdot \frac{x}{2} \, dx = 2 \]

Then, the variance is computed as:

• \[ V(X) = E(X^2) - [E(X)]^2 \]
• \[ = 2 - \left( \frac{4}{3} \right)^2 = \frac{2}{9} \]

The standard deviation \(\sigma_{X}\) is the square root of the variance:

• \[ \sigma_{X} = \frac{\sqrt{2}}{3} \]

This helps illustrate variability in the book checkout durations.

A Probability Density Function (PDF) provides the probabilities of a continuous random variable as a function of \(x\). Given a CDF, the PDF can be derived by differentiating the CDF.

In the problem, the PDF \(f(x)\) is the derivative of \(F(x) = \frac{x^2}{4}\) for \(0 \leq x < 2\). Differentiating leads to:

• \[ f(x) = \frac{d}{dx} \left( \frac{x^2}{4} \right) = \frac{x}{2} \]

This function describes the likelihood of a specific checkout time within the interval. The PDF is zero outside this range of \(0 \leq x < 2\), since the book cannot be checked out for less than zero or more than two hours.

Understanding the PDF is pivotal as it allows us to compute expected values and other statistical measures over the continuous range of the variable.