91Ó°ÊÓ

The Cumulative Distribution Function (CDF) is a fundamental concept in probability and statistics. It describes the probability that a random variable takes on a value less than or equal to a specific number. In the context of our exercise, the CDF is given as \( F(x) = \frac{1}{125} x^3 \) for \( 0 \leq x \leq 5 \).

The CDF increases as \( x \) increases, representing the cumulative probability up to \( x \). For example, by substituting a value within the range, such as \( x=3 \), we can find the probability that the assembly time is less than or equal to 3 minutes, which is \[ F(3) = \frac{1}{125} \times 3^3 = 0.216 \].

This means that there is a 21.6% chance the assembly time will be 3 minutes or less. The CDF is useful because it gives us a complete picture of the distribution of probabilities for different outcomes.

The Probability Density Function (PDF) is derived from the Cumulative Distribution Function (CDF) by taking its derivative. The PDF provides the relative likelihood of different outcomes. For our problem, the given CDF is \( F(x) = \frac{1}{125} x^3 \), and upon differentiation, we get the PDF as \( f(x) = \frac{d}{dx} F(x) = \frac{3}{125} x^2 \) for \( 0 \leq x \leq 5 \).

The PDF helps us understand the probability of the assembly time being within a specific interval. It is important to note that the probability for a single exact value in continuous distributions is technically zero; we use PDFs to find probabilities over intervals by integrating the PDF over that interval.

For instance, to find the probability that the assembly time is between 2 and 4 minutes, we would integrate the PDF from 2 to 4: \[ P(2 \leq X \leq 4) = \int_{2}^{4} \frac{3}{125} x^2 \ dx \].

The Expected Value, denoted \(E(X)\), is a measure of the central tendency or the 'average' of a random variable's possible outcomes. For our exercise, we calculated \(E(X)\) by integrating the product of \(x\) and the PDF: \( E(X) = \int_{0}^{5} x \left( \frac{3}{125} x^2 \right) \ dx = \frac{3}{125} \int_{0}^{5} x^3 \ dx \).

After solving the integral, we found \( E(X) = 3.75 \) minutes, meaning the average time to complete an assembly is 3.75 minutes. The expected value provides a single summary measure of the entire distribution of assembly times. Knowing the expected value helps in planning and anticipation of process duration.

Variance, denoted as \( \operatorname{Var}(X) \), measures the spread or dispersion of the random variable around the expected value. It is calculated as the expected value of the squared differences from the mean. For our problem, we computed the variance using: \( \operatorname{Var}(X) = E(X^2) - (E(X))^2 \).

First, we calculated \( E(X^2) \) by integrating \( x^2 \times PDF \): \( E(X^2) = \int_{0}^{5} x^2 \left( \frac{3}{125} x^2 \right) \ dx = \frac{3}{125} \int_{0}^{5} x^4 \ dx = 15 \).

Using \( E(X) = 3.75 \), we then found \( \operatorname{Var}(X) = 15 - (3.75)^2 = 0.9375 \).

A lower variance indicates that the assembly times are closely clustered around the mean, which in this case means most assembly times are close to 3.75 minutes.