Problem 12 - Algebra 2

Question

For Exercises \(9-14\) , use the following information. An exponential distribution has a mean of \(0.5 .\) Find each probability. $$ x<1 $$

Step-by-Step Solution

Answer

The probability \(P(X < 1) \approx 0.8647\).

Step 1: Understand the Exponential Distribution

An exponential distribution is a continuous probability distribution used to model the time or space between events in a Poisson process. The mean, also called the expected value, is given as 0.5 for this distribution. The probability density function for an exponential distribution with rate parameter \(\lambda\) is \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\).

Step 2: Determine the Rate Parameter \(\lambda\)

Since the mean of the exponential distribution is \(\frac{1}{\lambda}\), we have the mean as 0.5. Setting \(\frac{1}{\lambda} = 0.5\), we solve for \(\lambda\). Thus, \(\lambda = 2\).

Step 3: Use the Cumulative Distribution Function (CDF)

The cumulative distribution function for an exponential distribution is \(F(x) = 1 - e^{-\lambda x}\). This function gives the probability that a random variable \(X\) with an exponential distribution is less than or equal to \(x\).

Step 4: Calculate \(P(X < 1)\)

To find \(P(X < 1)\), we substitute \(x = 1\) and \(\lambda = 2\) into the CDF: \(P(X < 1) = F(1) = 1 - e^{-2 \cdot 1}\).

Step 5: Simplify the Expression

Calculate the exponent in the function: \(e^{-2} = \frac{1}{e^2}\). Thus, \(P(X < 1) = 1 - \frac{1}{e^2}\).

Step 6: Calculate the Probability

Approximating \(e^2\), which is roughly 7.389, the expression becomes \(P(X < 1) \approx 1 - \frac{1}{7.389}\). Thus, \(P(X < 1) \approx 1 - 0.1353 = 0.8647\).

Key Concepts

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a vital concept in probability and statistics. For an exponential distribution, the CDF provides a way to know the probability that a random variable is less than or equal to a certain value. This is useful for understanding the distribution and predicting outcomes.

In the case of the exponential distribution, the CDF is defined mathematically as:

• \( F(x) = 1 - e^{-\lambda x} \)

Here, \(e\) represents the base of the natural logarithm, and \(\lambda\) is the rate parameter. This function is quite intuitive once broken down:

- \( F(x) \) starts at zero when \(x\) is zero.
- As \(x\) increases, \(F(x)\) approaches 1 but never exceeds it.

This behavior shows that more of the probability is to the left of any given \(x\), meaning larger \(x\) values cover more of the distribution. For example, in this exercise, calculating the CDF at \(x = 1\) allowed us to determine the probability of the random variable being less than one.

Probability Density Function

The Probability Density Function (PDF) is another core concept of continuous distributions like the exponential distribution. Unlike the cumulative distribution function, which provides cumulative probabilities, the PDF gives the likelihood of the variable taking on a specific value, although for continuous variables, this is usually interpreted over an interval.

For the exponential distribution, the function is expressed as:

• \( f(x) = \lambda e^{-\lambda x} \) for \(x \geq 0\)

Here is how it works:

- \(\lambda\) is again the rate parameter, dictating the shape and scaling of the distribution.
- The exponential part \(e^{-\lambda x}\) decreases as \(x\) increases, showing diminishing likelihood of larger intervals.

This function gives a shifting sense of likelihood, where smaller \(x\) values are more probable in an exponential distribution. Thus, the PDF together with the CDF allows deeper insights into an event's timing, such as how quickly or slowly an event might happen.

Poisson Process

Understanding the Poisson process is key to contextualizing the exponential distribution. It applies to scenarios where events happen continuously and independently at a constant average rate. A classic example might be the number of phone calls at a call center.

The exponential distribution is particularly significant in a Poisson process because it describes the time between consecutive events. Simply put, if events are occurring randomly in time or space, the time until the next event is exponentially distributed. This supports practical applications, especially in areas such as reliability testing, survival analysis, and queueing theory.

The Poisson process has certain assumptions:

• Events occur independently.
• The average number of events in a fixed interval is constant.
• The probability of more than one event occurring in an infinitesimally small time frame is negligible.

This model and its relationship with the exponential distribution is powerful because it serves many areas of research and practical analysis.

Rate Parameter

The rate parameter, represented by \( \lambda \), greatly influences the character of an exponential distribution. It defines how quickly or slowly events are expected to occur within the model.

Key aspects of the rate parameter include:

• It is the inverse of the mean (\( \lambda = \frac{1}{\text{mean}} \)).
• The higher the value of \( \lambda \), the more rapidly events are expected to occur.
• A smaller \( \lambda \) suggests a slower rate of events.

In the original problem, with a mean of 0.5, the rate parameter \( \lambda \) was calculated as 2. This tells us that events, such as transactions or failures, are expected to happen fairly quickly. Understanding \( \lambda \) in real-world terms helps model the expected behavior or waiting times between consecutive occurrences, thus serving as a guiding parameter in predictive analytics and decision-making.