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The Cumulative Distribution Function (CDF) is a crucial concept when dealing with random variables and probability distributions. In the context of an exponential distribution, the CDF for a variable, say time spent on holiday shopping using an office computer, can be expressed as:\[ F(x; \lambda) = 1 - e^{-\lambda x} \]Here, \(x\) represents the amount of time, and \(\lambda\) is the rate parameter.
Using the CDF, we determine the probability that a random variable is less than or equal to a specific value. For instance, in our problem, using the CDF provided:- When seeking \(P(X \leq 5)\), we set the CDF equal to 0.53 to solve for the rate parameter \(\lambda\), yielding \(e^{-\lambda \cdot 5} = 0.47\) after rearranging.
This method demonstrates how the CDF is utilized to find probabilities and solve for important parameters like \(\lambda\). It essentially tells us the total probability of the outcomes that are less or equal to a particular point, a foundational aspect of understanding continuous distributions.

The rate parameter, denoted by \(\lambda\), is fundamental in defining an exponential distribution. It suggests how quickly or slowly events are expected to happen. A higher \(\lambda\) indicates events occurring rapidly, while a lower \(\lambda\) suggests a slower process.
For our example, by solving the equation \(1 - e^{-\lambda \cdot 5} = 0.53\), we determined \(\lambda\) to be approximately 0.14842.
The mean for an exponential distribution is simply the reciprocal of the rate parameter, \(\mu = \frac{1}{\lambda}\). Thus in our case:

• The mean \(\mu = \frac{1}{0.14842} \approx 6.74\).
• This value, when rounded, is closest to 6.6 hours, which is one of the options provided in the original exercise.

This relationship highlights why understanding the rate parameter is essential, as it directly influences the expected average number of events or occurrences.

Statistical probability is the backbone of predicting outcomes within specified limits or over certain intervals. With the exponential distribution, it allows us to determine the likelihood of a particular range of events.
For example, calculating the probability that a worker uses a computer for more than 10 hours involves utilizing the complement of the CDF:- The probability is given by \(P(X > 10) = 1 - P(X \leq 10) = e^{-\lambda \cdot 10}\).- Substituting our \(\lambda\), we get \(P(X > 10) \approx 0.2275\).
Similarly, finding the probability for a time range, like 4 to 8 hours, involves the difference in cumulative probabilities:

• \(P(4 < X < 8) = P(X \leq 8) - P(X \leq 4)\).
• This calculates to approximately 0.281, showing how statistical probability can pinpoint specific desired outcomes over ranges within the exponential spread.

Using statistical probability this way allows for comprehensive insights and risk assessments across varying scenarios.