Problem 58 - Calculus for Biology and Medicine

Question

Suppose the lifetime of a light bulb is exponentially distributed with mean 1 year. The light bulb is instantly replaced upon failure. What is the probability that, over a period of five years, at most five light bulbs are needed?

Step-by-Step Solution

Answer

The probability that at most five light bulbs are needed is approximately 0.6161.

Step 1: Define the parameters

The mean lifetime of a light bulb is given as 1 year. Since the lifetime is exponentially distributed, the rate parameter \( \lambda \) (which is the reciprocal of the mean) is 1. Therefore, \( \lambda = 1 \).

Step 2: Identify the distribution

Since the failure times are exponentially distributed with rate \( \lambda = 1 \), the number of failures (or light bulb replacements) in a given time period follows a Poisson distribution. For a period of \( t = 5 \) years, the Poisson distribution parameter is \( \lambda t = 5 \).

Step 3: Set up the Poisson probability

We need to calculate the probability of at most 5 failures (replacements) in 5 years, which is given by the sum of Poisson probabilities from 0 to 5, where the expected number of events is 5.

Step 4: Calculate the required probability

The probability of at most 5 light bulbs being needed, \( P(X \leq 5) \), is given by the sum: \[ P(X \leq 5) = \sum_{k=0}^{5} \frac{e^{-5} \cdot 5^k}{k!}. \] Calculate each term: - \( k = 0: \frac{e^{-5} \cdot 5^0}{0!} = e^{-5} \) - \( k = 1: \frac{e^{-5} \cdot 5^1}{1!} = 5e^{-5} \) - \( k = 2: \frac{e^{-5} \cdot 5^2}{2!} = \frac{25}{2} e^{-5} \) - \( k = 3: \frac{e^{-5} \cdot 5^3}{3!} = \frac{125}{6} e^{-5} \) - \( k = 4: \frac{e^{-5} \cdot 5^4}{4!} = \frac{625}{24} e^{-5} \) - \( k = 5: \frac{e^{-5} \cdot 5^5}{5!} = \frac{3125}{120} e^{-5} \) Adding these values up gives the total probability.

Step 5: Compute the final result

Calculate the numerical value for each term and sum them up: - \( e^{-5} \approx 0.00674 \)- \( 5e^{-5} \approx 0.0337 \)- \( \frac{25}{2} e^{-5} \approx 0.08425 \)- \( \frac{125}{6} e^{-5} \approx 0.14042 \)- \( \frac{625}{24} e^{-5} \approx 0.175525 \)- \( \frac{3125}{120} e^{-5} \approx 0.175525 \)Summing these values gives approximately \( 0.6161 \).

Key Concepts

Exponential Distribution

The lifetime of a light bulb in our scenario follows an exponential distribution. This type of distribution is used to model the time until the first occurrence of an event, such as failure or replacement, when events occur continuously and independently at a constant average rate.

Key characteristics include:

• Memoryless Property: The probability of failure in the future is not affected by how long the bulb has already lasted.
• Defined by the rate parameter \( \lambda \): For our case, the light bulb’s lifetime mean is 1 year, meaning \( \lambda = 1 \). This parameter is a reciprocal of the mean.

In our problem, this means that the expected number of failures, or replacements, of the light bulb in a given time follows another type of distribution. Since we know that the exponential distribution models the time between events, it directly connects to the Poisson distribution, which counts the number of events in a fixed interval of time.

Probability

In this exercise, we are interested in calculating the probability of replacing at most five light bulbs over a span of five years. This seems simple, but it touches on fundamental probability concepts.

Probability ranges from 0 to 1, representing an impossibility of an event to a certainty. In this case, our interest lies in a cumulative probability, i.e., the likelihood of observing up to a certain number of events.

To compute this probability, we use the Poisson probability mass function. It helps determine the likelihood of observing \( k \) events in an interval when we know \( \lambda \) (the expected rate of occurrence). Analysts sum the probabilities for all event counts from 0 up to 5 to find the total probability. This process ensures we cover all possibilities up to the event in question.

• The formula used: \[P(X \leq 5) = \sum_{k=0}^{5} \frac{e^{-5} \cdot 5^k}{k!},\]where \( e \) is the Euler’s number.
• Each term contributes to the overall probability of using a specific number of light bulbs.

Calculus

Calculus might not seem directly involved when solving this problem since it uses summation mostly. However, underlying these computations is the concept of limits and continuous change, aspects that calculus deeply explores.

Calculus provides the mathematical underpinning for exponential and Poisson distributions through derivative and integration concepts. For example, understanding how the Poisson process is derived from the exponential distribution integrates ideas of calculus. Here are a few insights on how these areas connect:

• Derivatives: Calculus is used to derive the exponential distribution function, which is foundational for computing continuous probabilities.
• Integrals: If we were handling more complex or non-linear models, integrals might help compute expected values over a continuous interval.

While cups of calculus aren't poured directly into the solution, students study its contributions through distributions, allowing predictive insights over continuous data.