Problem 23 - Calculus Concepts

Question

Suppose that \(g\) with input \(t\) is an exponential density function with \(k=2\). a. Find \(G,\) the corresponding cumulative distribution function. b. Use both \(g\) and \(G\) to calculate the probability that \(t \leq 0.35\) c. Use \(G\) to calculate the probability that \(t>0.86\).

Step-by-Step Solution

Answer

a. \(G(t) = 1 - e^{-2t}\). b. \(P(t \leq 0.35) \approx 0.5034\). c. \(P(t > 0.86) \approx 0.178\).

Step 1: Identify the Exponential Density Function

The probability density function (PDF) for an exponential distribution with rate parameter \(k\) is given by \(g(t) = ke^{-kt}\) for \(t \geq 0\). Here, the rate parameter \(k\) is given as 2, so the PDF is \(g(t) = 2e^{-2t}\).

Step 2: Determine the Cumulative Distribution Function (CDF)

The cumulative distribution function \(G(t)\) is found by integrating the PDF from 0 to \(t\). Thus, \(G(t) = \int_{0}^{t} 2e^{-2x}dx\). Solving this integral, we get \(G(t) = 1 - e^{-2t}\).

Step 3: Calculate Probability for Part (b)

To find \(P(t \leq 0.35)\) using \(G(t)\), substitute \(t = 0.35\) into the CDF: \(G(0.35) = 1 - e^{-2 \times 0.35}\). Compute this to get \(G(0.35) = 1 - e^{-0.7}\), which approximately equals 0.5034. This is the probability \(P(t \leq 0.35)\).

Step 4: Verify with PDF for Part (b)

Calculate \(P(t \leq 0.35)\) using \(g(t)\) by integrating the PDF from 0 to 0.35: \(\int_{0}^{0.35} 2e^{-2x}dx = 1 - e^{-0.7}\). This result verifies our CDF calculation: approximately 0.5034.

Step 5: Calculate Probability for Part (c)

To calculate \(P(t > 0.86)\) using \(G(t)\), recognize that \(P(t > 0.86) = 1 - P(t \leq 0.86)\). First, find \(G(0.86) = 1 - e^{-2 \times 0.86}\). Then, \(P(t > 0.86) = 1 - (1 - e^{-1.72}) = e^{-1.72}\), which is approximately 0.178.

Key Concepts

Exponential Distribution

The exponential distribution is a continuous probability distribution used to model the time until an event occurs. Typically, this involves scenarios like the time until a machine fails or the waiting time for a bus. It is characterized by a constant rate, meaning the event's occurrence is memoryless, which signifies that past events do not affect future probabilities. This makes it an essential tool in reliability testing and queuing theory.

Its probability density function (PDF) is given by:

• \( g(t) = ke^{-kt} \) for \( t \geq 0 \)

The parameter \(k\) here influences the shape and rate of the distribution. The higher the value of \(k\), the steeper the decline of the function, indicating a shorter expected wait time until the event occurs. The exponential distribution is integrally tied to the Poisson distribution, often representing the time between Poisson occurrences.

Cumulative Distribution Function

The cumulative distribution function (CDF) of an exponential distribution provides a way to calculate the probability that a random variable \(t\) will take a value less than or equal to a specific value. It accumulates probabilities from the left and reaches 1 as \(t\) approaches infinity.

The CDF is defined as:

• \( G(t) = p{0}^{t}g(x)dx \)

For the exponential distribution with \(g(t) = 2e^{-2t}\), integrating from 0 to \(t\) gives the cumulative function:

• \( G(t) = 1 - e^{-2t} \)

This equation highlights how the distribution's memoryless property operates over time. The cumulative function \(G(t)\) is particularly useful for calculating probabilities for ranges in continuous distributions.

Rate Parameter

The rate parameter \(k\) is a critical component in understanding the exponential distribution. It represents how quickly the event you're studying is expected to happen. In mathematical terms, it is the reciprocal of the mean; thus,

• \( ext{mean} = rac{1}{k} \)

A higher \(k\) value implies that the event is more frequent, leading to a steeper probability curve and shorter average time to the event. In the given problem, \(k\) is set to 2, meaning the average wait time for the event is 0.5 units of whatever time measurement is used.

The rate parameter is intrinsic to both the PDF and CDF, directly affecting the entire distribution's shape and behavior. It is vital in models where estimating time intervals is essential, like predicting life expectancy of systems or survival analysis.

Integration

Integration is a fundamental mathematical operation that helps determine the area under a curve. In the context of probability and statistics, integrating a Probability Density Function (PDF) over a specific range provides the probability of the random variable falling within that interval. For continuous distributions such as the exponential, this process is used to derive the cumulative distribution function (CDF).

To find the CDF \(G(t)\) of an exponential function, you integrate the PDF from zero to \(t\):

• \( G(t) = p{0}^{t} ke^{-kx}dx \)

In our example, evaluating this integral gives:

• \( G(t) = 1 - e^{-2t} \)

Essentially, integration transforms the instantaneous rate given by the PDF into a cumulative measure represented by the CDF. The process involves applying integration techniques, often requiring rules of calculus, to solve these mathematical expressions and translate them into probabilities.