91Ó°ÊÓ

A probability density function (PDF) is a function that describes the likelihood of a continuous random variable taking on a particular value. The area under the curve of the PDF over an interval represents the probability that the random variable falls within that interval. In this exercise, the given PDF is: \[f(x) = \frac{1}{72} x, \, 0 \leq x \leq 12,\]This means the probability density increases linearly from 0 to 12. The PDF must satisfy two conditions:

• The function must be non-negative for all values within the interval, i.e., \( f(x) \geq 0 \).
• The integral of the PDF over the interval must equal 1, ensuring the total probability is 1:

\[ \int_{0}^{12} f(x) \, dx = 1\]By integrating, you confirm that the PDF meets these conditions, ensuring it is a valid probability density function.

A continuous random variable can take any value within a given range. Unlike discrete random variables, which have distinct and separate values, continuous random variables have values that form a continuum. In this exercise, the time spent reading (in minutes) is a continuous random variable because it can take any value between 0 and 12. Understanding the behavior of a continuous random variable involves working with its probability density function. For instance, the expected value (mean) of the variable provides a measure of the central tendency. Finding this involves integrating over the possible values of the random variable, weighted by the PDF.The expected value of a continuous random variable with density function \( f(x) \) over the interval \( [a, b] \) is given by: \[ E(X) = \int_{a}^{b} x f(x) \, dx \]

Integral calculus is essential in probability theory, especially when working with continuous random variables. It allows us to compute areas under curves, which represent probabilities and expected values. Integrating a function involves finding the antiderivative and evaluating it within the given limits.In this exercise, we need to integrate to find the expected value: \[ E(X) = \int_{0}^{12} x f(x) \, dx \]By substituting the given PDF, the integral becomes: \[ E(X) = \frac{1}{72} \int_{0}^{12} x^2 \, dx \]We compute the integral of \( x^2 \) over 0 to 12: \[ \int_{0}^{12} x^2 \, dx = \left. \frac{x^3}{3} \right|_{0}^{12} = \frac{12^3}{3} - 0 = 576 \]Multiplying by the constant \( \frac{1}{72} \) gives us the expected value: \[ E(X) = \frac{1}{72} \times 576 = 8 \]Thus, we use integral calculus to determine that the average time spent reading the editorial page is 8 minutes.