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A probability density function (pdf) is a critical concept in statistics that describes the likelihood of a continuous random variable taking on a specific value. Unlike discrete probability distributions that use probabilities for individual outcomes, a pdf is used for continuous random variables where the exact probability of the random variable assuming a specific value is zero. Instead, we use a pdf to describe the probability of the variable being within a certain range of values.Some key characteristics of a pdf include:

• The function must be non-negative for all possible values.
• The integral of the pdf across the entire space is equal to 1, which ensures that the total probability is 1 (or 100%).

In our original problem, the pdf is given as:\[ f(r)=\left\{\begin{array}{cl} \frac{3}{4}\left[1-(10-r)^{2}\right] & 9 \leq r \leq 11 \ 0 & \text{otherwise} \end{array}\right.\]This implies that the radius \(R\) can vary between 9 and 11 meters, with different probabilities depending on the value of \(r\) within this range.

A continuous random variable is a type of variable that can take an infinite number of values within a given range. Unlike a discrete random variable, which can only assume distinct and separate values, a continuous random variable can take on any value in an interval. For continuous variables, probability calculations often require integration over the range of possible values to find the probability of the variable being in a specific interval.In the context of the exercise, the radius \(R\) of the circle is a continuous random variable. It is described by a probability density function, allowing it to assume any real value between 9 and 11 meters. This randomness stems from variations in the actual measured radius compared to the intended 10 meters.Importantly, for continuous random variables, the probability of the variable assuming any specific value itself is zero. Instead, we calculate probabilities over intervals, often involving integration. For instance, to calculate the expected value, which is a type of average value of the random variable, we integrate using the pdf over its range.

The formula for the area of a circle is essential in this exercise because the expected area is what we're calculating. The area \(A\) of a circle is given by the formula:\[ A = \pi r^2 \]This basic geometric formula provides the relationship between the radius and the area of the circle.In our problem, because the radius \(R\) is not fixed and can vary based on a probability distribution, calculating the area is more complex than merely inserting a radius into the formula. We use the expected value of the area, \(E(\pi R^2)\), by integrating over the possible values of \(R\) given by the pdf.This involves setting up an integration where the function inside the integration is the area formula itself weighted by the probability of each radius value. By solving this integration, we find the expected, or mean, area accounting for all possible radii \(R\) and their likelihood.

Integral calculus is a branch of mathematics used to calculate areas, volumes, and accumulations, among other things. It plays a vital role in finding the expected values for continuous random variables, like in our original problem.Integrals allow us to calculate the area under curves, such as the area under a probability density function, to find total probabilities or expected values. In this exercise, we use integral calculus to find the expected value of \(\pi R^2\), which represents the expected area of the circular region.The integral used in this example is:\[ \int_{9}^{11} \pi r^2 \cdot \frac{3}{4}\left(1 - (10-r)^2\right) \, dr \]This integral calculates the area under the curve of the function described, over the interval from 9 to 11. It helps account for the probability-adjusted area calculations, giving us a way to express the likely average area of the sampling region with the random radius \(R\).