Problem 57 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

(a) The probability is approximately 0.4866. (b) The probability is approximately 0.3056.

Step 1: Understanding the Exponential Distribution

The exponential distribution is defined by the probability density function (PDF) \( f(t) = \lambda e^{-\lambda t} \), where \( \lambda \) is the rate parameter. The mean of an exponentially distributed random variable \( X \) is given by \( \frac{1}{\lambda} \). In this case, the mean is 3 years, which means \( \frac{1}{\lambda} = 3 \), so \( \lambda = \frac{1}{3} \).

Step 2: Calculate Probability for Part (a)

For part (a), we want the probability that the device fails after 2 years. This means calculating the cumulative distribution function (CDF), given by \( F(t) = 1 - e^{-\lambda t} \). Substituting \( \lambda = \frac{1}{3} \) and \( t = 2 \), we have: \[ P(T > 2) = 1 - e^{-\frac{1}{3} \cdot 2} = e^{-\frac{2}{3}}. \] Numerically calculate \( e^{-\frac{2}{3}} \) to get the result.

Step 3: Understanding Poisson Distribution for Part (b)

When events occur independently and continuously at a constant average rate, the number of events in a fixed interval follows a Poisson distribution. Here, the average number of failures in 5 years is \( \mu = \lambda \cdot 5 = \frac{1}{3} \cdot 5 = \frac{5}{3} \).

Step 4: Calculate Probability for Part (b)

For part (b), we want the probability of exactly one failure in 5 years. Using the Poisson probability mass function (PMF): \[ P(N = 1) = \frac{(\frac{5}{3})^1 \cdot e^{-\frac{5}{3}}}{1!} = \frac{5}{3} \cdot e^{-\frac{5}{3}}. \] Compute the expression \( \frac{5}{3} \cdot e^{-\frac{5}{3}} \) to obtain the probability.

Key Concepts

Probability Density Function

The probability density function, or PDF, characterizes the likelihood of a continuous random variable taking on a particular value. For the exponential distribution, which we are examining, the PDF is expressed as: \[f(t) = \lambda e^{-\lambda t}\]This formula tells us how the probability is spread out over different values of the random variable. The parameter \( \lambda \), known as the rate parameter, controls how fast the exponential function decays.
- The higher the value of \( \lambda \), the steeper the curve, resulting in a higher chance that the event occurs quickly.- The shape of the exponential PDF is always a declining curve, indicating that the probability decreases over time.
The PDF applies to all non-negative values of time \( t \), showing that as time progresses, the probability density gradually decreases. This suits scenarios where events occur continuously and independently.

Cumulative Distribution Function

In contrast to the PDF, the cumulative distribution function (CDF) helps determine the probability that the random variable will be less than or equal to a certain value. For exponential distribution, the CDF is formulated as:\[F(t) = 1 - e^{-\lambda t}\]Using this formula, you can find the probability that an event, like the failure of a device, will occur after a specific time.

• The value \( F(t) \) increases as \( t \) increases, never exceeding 1, confirming the total probability of all possible outcomes is 1.
• At time \( t = 0 \), the CDF is 0, indicating no time has elapsed, so the probability is 0 that the event has already occurred.

Remember, CDF complements PDF by accumulating probabilities over time, offering a broader view of how likely it is for the event to happen within a specified duration.

Poisson Distribution

The Poisson distribution is essential for scenarios where we count occurrences of events within a fixed interval, given a constant mean rate. When linked to the exponential distribution, it models the number of events occurring in a time period.
For example, if we consider a device that fails at an average rate \( \lambda \) every year, the number of failures in the timeframe follows a Poisson distribution.

• The mean of a Poisson distribution, represented by \( \mu \), can be calculated as \( \lambda \times \text{time} \).
• It characterizes discrete data which means it concerns counts of events that happen at distinct points in time.

When applying Poisson, the focus is on how many times an event might happen rather than when it happens, perfect for tallying occurrences within specified periods.

Rate Parameter

The rate parameter, \( \lambda \), is a crucial component in both exponential and Poisson distributions. In essence, it denotes how frequently events are expected to take place over time.

• For the exponential distribution, \( \lambda \) determines the rapidity at which the event timing declines. A higher \( \lambda \) results in more frequent events.
• In the Poisson distribution, \( \lambda \) helps in defining the expected number of occurrences in a given time period.

By understanding \( \lambda \), one gains insight into the pace and intensity of the processes being analyzed. This parameter establishes a bridge linking time-dependent phenomena, ensuring models reflect the true nature of the occurrences.