91Ó°ÊÓ

The Cumulative Distribution Function (CDF) is a fundamental concept in probability distributions. It represents the probability that a random variable, say \( X \), will be less than or equal to a certain value \( x \). For any value \( x \), the CDF is expressed as \( F(x) = P(X \leq x) \). This function helps us understand how probabilities accumulate over a range of possible values.In our problem, the task completion time \( X \) has a CDF, \( F(x) \), which is piecewise defined across different intervals:

• For \( x < 0 \), \( F(x) = 0 \), indicating no probability mass below zero since the task cannot take negative time.
• For \( 0 \leq x < 1 \), \( F(x) = \frac{x^3}{3} \), showing a cubic accumulation of probability.
• For \( 1 \leq x \leq \frac{7}{3} \), it involves a more complex calculation: \( F(x) = 1-\frac{1}{2}((\frac{7}{3}-x)(\frac{7}{4}-\frac{3}{4}x)) \).
• Lastly, for \( x \geq \frac{7}{3} \), \( F(x) = 1 \), indicating all possible probability is accumulated by this point.

Understanding the CDF's form over different intervals is crucial for finding other properties like the PDF and expectation.

The Probability Density Function (PDF) is derived by differentiating the Cumulative Distribution Function (CDF). The PDF provides the likelihood of the random variable taking on a specific value. It's crucial to remember that the PDF itself is not a probability but rather a density that can be integrated over an interval to find probabilities.To find the PDF, we differentiate the given CDF for each defined interval:

• In the interval \( 0 \leq x < 1 \), where \( F(x) = \frac{x^3}{3} \), differentiating gives us \( f(x) = x^2 \). This represents a quadratic increase in probability density as \( x \) approaches 1.
• For \( 1 \leq x \leq \frac{7}{3} \), \( F(x) \) needs to be differentiated: \( f(x) = \frac{1}{2}(\frac{9}{4} - \frac{3}{2}x) \). Here, the density begins to tapers off linearly.
• For \( x < 0 \) and \( x \geq \frac{7}{3} \), \( f(x) = 0 \), reflecting no probability density in these regions.

The PDF is useful when you need to compute probabilities over specific intervals of the random variable \( X \), by integrating the function over the desired interval.

The expectation or expected value of a continuous random variable gives us the long-run average outcome of a given random phenomenon. Conceptually, it is the center or "balance point" of the probability distribution.In mathematical terms, for a continuous random variable with PDF \( f(x) \), the expected value \( E(X) \) is computed as:\[ E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \]In practice, we often only integrate over the domain where \( f(x) \) is not zero.For the provided exercise, \( E(X) \) is calculated using two integrals corresponding to non-zero portions of \( f(x) \):

• From \( 0 \) to \( 1 \): \( \int_{0}^{1} x \cdot x^2 \, dx \).
• From \( 1 \) to \( \frac{7}{3} \): \( \int_{1}^{\frac{7}{3}} x \cdot \frac{1}{2}(\frac{9}{4}-\frac{3}{2}x) \, dx \).

Solving these integrals gives \( E(X) = \frac{27}{40} \), which provides insight into the expected completion time of the task. The expected value not only informs about the average case but also helps in planning and decision-making processes. By understanding these calculations, students can apply the concept to similar continuous distributions.