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In probability and statistics, the expected value of a random variable is a core concept that represents the average outcome if an experiment was repeated a large number of times. For a random variable \( t \) following an exponential distribution, calculating the expected value is straightforward. The formula for the expected value \( E(t) \) in an exponential distribution with a rate parameter \( \lambda \) is given by:\[ E(t) = \frac{1}{\lambda} \]This formula shows how the rate parameter inversely affects the expected outcome. For example, in our exercise, the rate parameter \( \lambda \) is 3, so:\[ E(t) = \frac{1}{3} \]This means that, on average, the value of \( t \) will be \( \frac{1}{3} \) in the context of this exponential distribution. This expected value provides a central tendency, giving us a sense of the 'long-term' average result.

The standard deviation is a measure of how spread out the values of a random variable are around the mean. For an exponential distribution, once we've determined the variance, calculating the standard deviation becomes a simple task. Firstly, we need the formula for the variance for an exponential distribution with rate \( \lambda \):\[ \text{Var}(t) = \frac{1}{\lambda^2} \]For our particular problem, where \( \lambda = 3 \), we plug it into the formula:\[ \text{Var}(t) = \frac{1}{3^2} = \frac{1}{9} \]Since standard deviation is the square root of variance, the calculation becomes:\[ \sigma = \sqrt{\frac{1}{9}} = \frac{1}{3} \]The standard deviation tells us that most of the data from this exponential distribution will fall within \( \frac{1}{3} \) units from the mean. Understanding this can help one comprehend the dispersion of outcomes around the expected value.

A probability density function (PDF) defines the likelihood of a continuous random variable taking on a particular value. For the exponential distribution, the PDF is expressed as:\[ H(t) = \lambda e^{-\lambda t} \quad \text{for}\; t \geq 0 \]In our case, the PDF is given by \( H(t) = 3e^{-3t} \) where the rate parameter \( \lambda \) is 3. The PDF helps us to understand how probabilities are distributed over the values of \( t \). Some critical points about the PDF:

• The area under the curve of the PDF over its entire range is always equal to 1, indicating total probability.
• The shape and position of the curve tell us how probabilities are assigned to different outcomes.
• For an exponential distribution, the function decreases exponentially, highlighting that higher values of \( t \) are less probable.

The exponential nature of the PDF makes it particularly useful in modeling time until the next event occurs, like the time between arrivals in a queue or the lifespan of a decaying substance. Understanding the PDF helps in interpreting how likely different scenarios are to happen.