91Ó°ÊÓ

A continuous random variable, like the one described in the exercise, is a type of variable that can take an infinite number of possible values within a given range. In this context, the random variable \(X\) represents the amount of time a book is checked out from the library.

Unlike discrete random variables, which have specific, distinct values, continuous random variables have a continuum of possibilities. This means that within any interval, no matter how small, there are infinite possibilities.

For example, \(X\) could be 1.3 hours, 1.3001 hours, or 1.30000001 hours. Therefore, to calculate probabilities, we rely on calculus, specifically integration, to find the area under the curve of the density function. This area gives us the likelihood of \(X\) falling within a specific range.

The density function of a continuous random variable describes the likelihood of the variable adopting a specific value, or range of values. In this exercise, the density function is defined piecewise, as \(f(x) = 0.5x\) for \(0 \leq x \leq 2\) and \(f(x) = 0\) elsewhere.

A density function differs from a probability mass function, which is used for discrete variables. For continuous variables, it's not about individual probabilities since the probability of the random variable being exactly any single value is zero. Instead, the focus is on ranges or intervals.

The density function must satisfy two conditions:

• The function value is non-negative for all \(x\).
• The integral of the function over all possible values equals 1.

This ensures that it accurately represents the probability distribution of the continuous random variable.

Integral calculation is crucial to finding probabilities associated with continuous random variables. It involves finding the area under the curve of the density function over a specified interval. This process is akin to summing the infinitesimally small probabilities across a continuum of values.

The exercise requires integrating the density function, \(f(x) = 0.5x\), over specific intervals to find the probabilities. For problem (a), this involves calculating \(\int_{0}^{1} 0.5x \, dx\), which results in \(0.25\). Similarly, for problem (b), the calculation \(\int_{0.5}^{1.5} 0.5x \, dx\) yields \(0.5\).

Remember:

• Integral bounds define the interval you are evaluating.
• The integral symbol \(\int\) denotes that you're calculating the area under the curve.
• Evaluate the antiderivative at the upper and lower bounds to find the probability.

Integrals are indispensable in the realm of continuous probability distributions as they convert density functions into actionable probabilities.

A piecewise function is one that is defined by different expressions depending on the interval of the independent variable. In this problem, the density function is expressed piecewise as it changes its behavior outside the interval \([0, 2]\).

Such functions are essential in modeling situations where different rules apply over different ranges. Here, \(f(x) = 0.5x\) for \(0 \leq x \leq 2\), which indicates the function's behavior within the valid range for borrowing time. Outside this interval, \(f(x) = 0\), meaning there's no probability as the book can't be borrowed for less than 0 or more than 2 hours.

In practice:

• Understand each "piece" and the range it applies to.
• Apply the appropriate piece of the function while integrating over specific intervals.
• Ensure transitions between pieces are smooth and satisfy the condition of total probability being 1.

Piecewise functions offer a flexible way to model real-world problems with varying conditions.