Problem 61 - Calculus for Biology and Medicine

Question

Suppose the lifetime of a printer is exponentially distributed with parameter \(\lambda=0.2 /\) year. (a) What is the expected lifetime? (b) The median lifetime is defined as the age \(x_{m}\) at which the probability of not having died by age \(x_{m}\) is \(0.5\). Find \(x_{m}\).

Step-by-Step Solution

Answer

Expected lifetime: 5 years; Median lifetime: 3.47 years.

Step 1: Understand the Exponential Distribution

The exponential distribution is used to model the time between events in a Poisson process. It is characterized by the parameter \(\lambda\), which represents the rate of occurrence of events. The probability density function of an exponential distribution is given by \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\).

Step 2: Calculate Expected Lifetime

The expected value (mean) of an exponentially distributed random variable with parameter \(\lambda\) is given by \(E(X) = \frac{1}{\lambda}\). Substituting the given \(\lambda = 0.2\), we get:\[E(X) = \frac{1}{0.2} = 5 \, \text{years}.\]

Step 3: Define Median Lifetime Condition

The median lifetime \(x_m\) is defined such that the cumulative distribution function \(F(x_m)\) satisfies \(P(X > x_m) = 0.5\). For the exponential distribution, \(P(X > x_m) = 1 - F(x_m)\) where \(F(x_m) = 1 - e^{-\lambda x_m}\).

Step 4: Set Up the Equation for Median Lifetime

We need to solve the equation: \[ P(X > x_m) = 0.5. \] This implies: \[ 1 - F(x_m) = 0.5 \] which further leads to: \[ e^{-\lambda x_m} = 0.5. \]

Step 5: Solve for \(x_m\)

Take the natural logarithm on both sides of the equation \(e^{-\lambda x_m} = 0.5\): \[ -\lambda x_m = \ln(0.5) \] \[ x_m = -\frac{\ln(0.5)}{0.2}. \]Calculate the value to find:\[ x_m \approx -\frac{-0.6931}{0.2} \approx 3.4655\, \text{years}.\]

Step 6: Summary of Solutions

The expected lifetime of the printer is 5 years, and the median lifetime, \(x_m\), where there is a 50% chance of failure, is approximately 3.47 years.

Key Concepts

Expected Lifetime

The expected lifetime of a random variable modeled by the exponential distribution is a very important concept in probability and statistics. When we talk about expected lifetime, we mean the average time it will take for an event to occur for the first time. For an exponential distribution characterized by the rate parameter \( \lambda \), the expected lifetime can be calculated with a simple formula. The formula is given by:

• \( E(X) = \frac{1}{\lambda} \)

This formula shows that the expected lifetime is the reciprocal of the rate parameter. In practical terms, it tells us how long we can expect an event to "last" on average before it occurs. For example, if a printer's lifetime is exponentially distributed with a rate \( \lambda = 0.2 \)/year, the expected lifetime is:

• \( E(X) = \frac{1}{0.2} = 5 \) years

This means, on average, we expect the printer to work for 5 years before it fails. It is a key prediction that helps in planning and reliability studies.

Median Lifetime

The median lifetime is a statistic that gives us insight into the lifespan of an object in a probabilistic sense. Specifically, it tells us the time at which we have a 50% chance of an item failing. For an exponential distribution, the median lifetime, \( x_m \), can be found using a condition on the cumulative distribution function.Unlike the mean, which gives an average, the median tells us the 'middle point,' where half of the items have failed. To find the median \( x_m \), we set \( P(X > x_m) = 0.5 \). This condition corresponds to the expression:

• \( e^{-\lambda x_m} = 0.5 \)

By solving the equation:

• \( x_m = -\frac{\ln(0.5)}{\lambda} \)

we find that for \( \lambda = 0.2 \)/year, the median lifetime is approximately 3.47 years. It is less than the expected lifetime because the exponential distribution is skewed, meaning more observations are on the lower side.
The median gives a robust statistic that is useful for understanding the distribution of lifetimes in a population.

Probability Density Function

A probability density function (PDF) provides a way to describe the likelihood of a random variable. For a continuous random variable like those in exponential distributions, the PDF describes the relative likelihood of different outcomes.In an exponential distribution with rate parameter \( \lambda \), the PDF is represented as:

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

This function describes the likelihood of the time until the next event occurring. The PDF is high when an event is very likely to occur soon and decreases exponentially as time increases. This means that events are more likely to occur closer to the starting point and less so as time progresses.Understanding the PDF helps in predicting when an event is most likely to happen and in visualizing how the likelihood of events decrease over time. It plays a crucial role in statistical inference and decision-making processes.

Poisson Process

The Poisson process is a fundamental concept in the study of random events that occur independently over time. It serves as the basis for the exponential distribution and is characterized by a consistent average rate of occurrence.What makes a Poisson process unique is:

• Events occur independently of each other.
• The average rate, \( \lambda \), is constant over time.
• Two events cannot occur at the exact same time.

If the occurrence of events follows a Poisson process, the time between each event is exponentially distributed. The exponential distribution then provides the framework for understanding the intervals between those events.This concept is essential in a multitude of fields including queuing theory, telecommunications, and reliability engineering. It helps in determining when events are expected to occur, assisting in optimizing processes and improving efficiency.