91Ó°ÊÓ

The cumulative distribution function (CDF) is a fundamental concept in probability and statistics. It tells us the probability that a random variable takes on a value less than or equal to a particular value. In our context, the CDF for an exponentially distributed random variable, such as the life length of a component \(Y\), is used to find the distribution function of another variable \(X\), which represents the time until failure or replacement.

In the problem, \(X = \min(Y, 200)\), meaning it's the lesser of \(Y\) or 200 hours. The component could fail (\(Y < 200\)) or be replaced at 200 hours. For \(0 \leq x < 200\), \(F_X(x) = P(X \leq x) = P(Y \leq x) = 1 - e^{-x/100}\). This equation comes from the exponential distribution, capturing the probability up to time \(x\).

For \(x \geq 200\), the CDF becomes \(F_X(x) = 1\), indicating that by 200 hours, the component is definitely replaced or failed. Thus, the CDF provides a complete picture of how the time \(X\) is distributed.

Expectation of a random variable is a vital tool to measure its 'average' or 'mean' value. It summarizes the central tendency of its probability distribution. In this exercise, we seek \(E(X)\), the expected life span of the component.

For the time of usage \(X\), expectation needs to consider both situations: the component fails before 200 hours, and it is replaced at 200 hours. We calculate \(E(X)\) by weighing these scenarios. For \(Y < 200\), the expected time is \(E(Y \mid Y < 200)\), reflecting the mean life span conditional on the event \(Y < 200\). On the other side, if \(Y \geq 200\), it's simply 200 hours.

This weighted average requires calculating probabilities and integrals, and involves steps such as finding \( P(Y < 200) = 1 - e^{-2} \) and specific expectations conditional on \(Y < 200\). The result for \(E(X)\) balances both scenarios, yielding approximately 86.47 hours.

Mean of an exponential distribution, often called its mean life, provides the average time between events in a process where events occur continuously and independently at a constant rate. For an exponential distribution with rate \( \lambda \), the mean \( \mu \) is simply \( \frac{1}{\lambda} \).

In this problem, the component's life length has a mean of 100 hours, thus \( \lambda = \frac{1}{100} \). The interpretation is that, on average, the component is expected to last 100 hours before failure, assuming no intervention like replacement based on time limit. This mean is crucial for determining other properties like the CDF and expectation.

The nature of exponential distributions simplifies many calculations. Its memoryless property is useful because it implies that the probability of failure in future intervals depends only on the length of those intervals, not on when the observation or component life began.