Problem 57 - Calculus for Biology and Medicine

Question

Suppose the lifetime of a lightbulb is exponentially distributed with mean 3 years. The lightbulb is instantly replaced upon failure. (a) Find the probability that the lightbulb will have failed after two years. (b) What is the probability that, over a period of five years, the lightbulb was replaced only once?

Step-by-Step Solution

Answer

(a) Probability that the bulb fails in 2 years: \( 1 - e^{-2/3} \). (b) Probability of 1 replacement in 5 years: \( \frac{5}{3} e^{-5/3} \).

Step 1: Understand the mean of exponential distribution

Exponential distribution is defined by the parameter \( \lambda \), the rate of decay. The mean \( \mu \) of an exponential distribution is given by \( \mu = \frac{1}{\lambda} \). If the mean is 3 years, \( \lambda = \frac{1}{3} \).

Step 2: Find the probability of failure in two years

For an exponential distribution, the probability of failure before time \( t \) is given by \( 1 - e^{-\lambda t} \). Substituting \( \lambda = \frac{1}{3} \) and \( t = 2 \), the probability: \[ P(T < 2) = 1 - e^{-\frac{1}{3} \times 2} = 1 - e^{-\frac{2}{3}} \].

Step 3: Understand failure replacements over a time period

Over five years, we need to find the number of replacements. In an exponential distribution with rate \( \lambda \), the number of events (failures) in time \( t \) follows a Poisson distribution with parameter \( \lambda t \).

Step 4: Use Poisson distribution for five years analysis

The parameter \( \lambda t \) is \( \frac{1}{3} \times 5 = \frac{5}{3} \). The probability a Poisson random variable with parameter \( \frac{5}{3} \) equals 1 is: \[ P(N = 1) = \frac{{\left(\frac{5}{3}\right)^1 e^{-\frac{5}{3}}}}{1!} = \frac{5}{3} e^{-\frac{5}{3}} \].

Key Concepts

Poisson Distribution

The Poisson Distribution is a probability distribution that expresses the likelihood of a given number of events occurring in a fixed interval of time or space. This distribution is commonly used when these events occur with a known constant mean rate and independently of the time since the last event. In the context of the lightbulb problem, the events we are considering are the failures of the lightbulb.

• Here, the rate of events (or failures) is represented by the parameter \( \lambda t \), which is the product of the decay rate \( \lambda \) and the time period \( t \).
• For example, in our exercise, when we calculate for five years, the parameter becomes \( \frac{5}{3} \).

The Poisson distribution is particularly useful in situations where you're interested in finding out how often an event is likely to happen over a continuous interval, making it ideal for predicting the number of failures or arrivals, such as how many lightbulbs fail over a period of time.

Probability Calculation

Probability Calculations in the exponential and Poisson contexts typically use specific formulas to determine the likelihood of events over a time frame. For the exponential distribution:

• The probability that an event (like a lightbulb failing) happens before a certain time is \( P(T < t) = 1 - e^{-\lambda t} \).
• This formula is essential for part (a) of our problem. By putting \( \lambda = \frac{1}{3} \) and \( t = 2 \), you calculate the probability that the bulb fails in the first two years.

When it comes to Poisson Calculations:

• The formula for finding the probability that the lightbulb is replaced once in five years is given by the Poisson probability mass function: \( P(N = k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!} \), where \( k \) is the number of occurrences.
• For our problem, \( k = 1 \) and \( \lambda t = \frac{5}{3} \).

These calculations allow us to understand the likelihood of different scenarios occurring and are fundamental in statistical analysis and data science.

Mathematical Expectation

In the realm of probability and statistics, the Mathematical Expectation, or expected value, is a fundamental concept that provides the average or expected outcome of a random event. For an exponential distribution:

• The mean (or expected value) \( \mu \) of an exponential distribution is the average time you'd expect to wait for an event to occur, given by \( \mu = \frac{1}{\lambda} \).
• In our example, where \( \lambda = \frac{1}{3} \), the mean becomes 3 years. This tells us, on average, how long a lightbulb is expected to last before failing.

Understanding the mathematical expectation provides insights into long-term trends and can inform decision-making in practical applications, such as inventory and supply chain management.

Failure Rate

The Failure Rate, often symbolized by \( \lambda \) in an exponential distribution, is a crucial concept in reliability engineering and survival analysis. It describes how quickly failures are likely to happen over time:

• In exponential models, it is constant over time, signifying that each moment has the same likelihood of failure.
• This rate (\( \lambda \)) is inversely related to the mean (or expected lifetime), helping establish a straightforward relationship between how often we expect systems to fail and how long we anticipate they last.

For our lightbulb problem, knowing \( \lambda = \frac{1}{3} \) tells us that each year, every lightbulb has a manageable and consistent failure rate. Such knowledge is vital for planning maintenance and anticipating replacements, ensuring systems remain operational.