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Understanding the expected value is crucial when working with random variables, such as time to complete tasks. In the given exercise, the worker's times to complete three jobs are modeled as independent exponential random variables with mean 1, noted as \(T_{1}\), \(T_{2}\), and \(T_{3}\). The expected value, often denoted as \(E[X]\), is a fundamental concept in probability that represents the average outcome if an experiment is repeated a large number of times.

The calculation for the expected value of \(X\), which is the sum of times for the three jobs (\(C_{i}\)), is straightforward due to the identical distribution of the jobs' completion times. Since each job has a mean (and thus an expected value) of 1, the expected value for the total time \(X\) can be expressed as:
\[E[T_1] = E[T_2] = E[T_3] = 1\]
Summing these gives \(E[X] = 3\). This result underscores that the average time for the worker to finish all three jobs is three times the average time taken to complete one job.

The linearity of expectation is a powerful property that greatly simplifies the work with expected values, especially when dealing with sums of random variables. It states that the expected value of the sum of random variables is the sum of their expected values, irrespective of whether the random variables are independent or not.

In the context of the exercise, linearity of expectation allows us to find the expected value of \(X\), the total completion time for the three jobs, by simply adding up the expected values of each individual job's completion time. Using the property, we get:
\[E[X] = E[T_1 + T_2 + T_3] = E[T_1] + E[T_2] + E[T_3]\]
Since we know from the previous section that \(E[T_1] = E[T_2] = E[T_3] = 1\), we can conclude that \(E[X] = 3\). This is a neat example of how the linearity of expectation can make calculations more manageable and is a fundamental tool in analyzing the behavior of random variables.

Variance measures the spread or dispersion of a set of values. For random variables, it quantifies how much the values of the variable deviate from the expected value. In the exercise scenario, we're asked to find the variance of the sum of completion times. The variance of an exponential distribution with mean 1 is equal to the square of the mean. Therefore, for each job's time which we denote as \(T_i\), we have:
\[Var(T_i) = 1^2 = 1\]
The property that comes into play for calculating the variance of \(X\), \(Var(X)\), is specific to independent random variables: the variance of the sum is the sum of their variances. Thus, we can find \(Var(X)\) by summing the variances of \(T_1\), \(T_2\), and \(T_3\):
\[Var(X) = Var(T_1 + T_2 + T_3) = Var(T_1) + Var(T_2) + Var(T_3) = 1 + 1 + 1 = 3\]
This demonstrates that the total variability in the sum of the workers' completion times is three times the variability of a single job's time. The variance is a vital concept in understanding the reliability and consistency of processes, like how predictably a worker finishes tasks.