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A continuous random variable is a type of variable that can take an infinite number of possible values within a given range. Unlike discrete random variables that have a countable set of outcomes like 1, 2, 3, etc., continuous random variables can take on any value within a given interval. This makes them particularly useful for modeling measurements like weight, height, and temperature, which can vary continuously. For a continuous random variable, such as the one described in our exercise, the probability that the random variable takes on a specific value is zero. Instead, we focus on the probability that it falls within a particular range. This is where the concept of the probability density function (PDF) comes into play, which helps us calculate these probabilities effectively.

The probability density function (PDF) is a fundamental concept when dealing with continuous random variables. It provides a way to describe the likelihood of different outcomes over an interval. For a continuous random variable like \(X\), the PDF, denoted by \(f_{X}(x)\), is defined such that the area under the curve between two points gives the probability that the random variable falls within that range.In our exercise, the PDF for \(X\) is \(f_{X}(x) = \frac{x}{18}\) for \(0 \leq x \leq 6\). This function shows how the values of \(X\) are distributed over the interval from 0 to 6. The PDF needs to satisfy two main properties:

• It must be non-negative for all values of \(x\).
• The total area under the PDF curve must equal 1, representing the entire probability space.

Understanding how to work with the PDF is essential for transforming continuous random variables, as it allows us to find new distributions when we apply transformations, such as \(Y = 2X + 10\) in the exercise.

The expected value, often referred to as the mean, is a measure of the central tendency of a probability distribution. For a continuous random variable like \(X\), the expected value \(E[X]\) is computed by integrating over the entire range of the variable, weighted by its probability density function. It represents the average outcome we would expect if we repeated an experiment a large number of times.Mathematically, the expected value is calculated as:\[ E[X] = \int_{- \infty}^{\infty} x \cdot f_{X}(x) \, dx \]However, because \(f_{X}(x)\) is nonzero only over the interval from 0 to 6 in our example, we actually compute:\[ E[X] = \int_{0}^{6} x \cdot \frac{x}{18} \, dx \]When working with transformations, like \(Y = 2X + 10\), the expected value of \(Y\) can be easily determined using properties of linear transformations. The relationship \(E[Y] = 2E[X] + 10\) showcases the linear property which allows us to calculate the expected value of \(Y\) without explicitly finding its pdf. This saves time and simplifies calculations in many practical applications.