Problem 59 - Calculus for Biology and Medicine

Question

Suppose the lifetime of a technical device is exponentially distributed with mean five years. (a) Find the probability that the device will have failed after three years. (b) Given that the device has worked for six years, find the probability that it will work for another year.

Step-by-Step Solution

Answer

(a) 0.4512; (b) 0.8187

Step 1: Understand the Exponential Distribution

The exponential distribution is often used to model the time between events. It is defined by a rate parameter \( \lambda \), which is the reciprocal of the mean. Since the mean is given as 5 years, we find \( \lambda \) by taking the reciprocal: \( \lambda = \frac{1}{5} = 0.2 \).

Step 2: Find the Probability for Part (a)

To find the probability that the device will have failed after three years, use the survival function: \( P(T > t) = e^{-\lambda t} \). Find the complementary probability for failure: \( P(T \leq 3) = 1 - e^{-0.2 \times 3} = 1 - e^{-0.6} \). Calculate \( 1 - e^{-0.6} \).

Step 3: Calculate Probability for Exp(-0.6)

Calculate \( e^{-0.6} \). Using a calculator, \( e^{-0.6} \approx 0.5488 \). Thus, the probability of failure after 3 years is \( 1 - 0.5488 = 0.4512 \).

Step 4: Use Memoryless Property for Part (b)

Exponential distribution has the memoryless property, meaning: \( P(T > 7 | T > 6) = P(T > 1) \). We find the probability that the device works for another year given it has already worked 6 years: \( P(T > 1) = e^{-0.2 \times 1} = e^{-0.2} \).

Step 5: Calculate Probability for Exp(-0.2)

Calculate \( e^{-0.2} \). Using a calculator, \( e^{-0.2} \approx 0.8187 \). Thus, the probability that it will work for another year given it has already worked for six years is \( 0.8187 \).

Key Concepts

Mean Lifetime

The mean lifetime of a device or system that follows the exponential distribution is a central concept. This mean lifetime is the average time until failure, which in our exercise is given as 5 years. The exponential distribution is parameterized by the rate parameter \( \lambda \), which is the reciprocal of the mean lifetime. Therefore, in this context, \( \lambda = \frac{1}{5} = 0.2 \). This means that on average, one device fails every 5 years, but the actual lifetime of each device varies.The understanding of mean lifetime helps us calculate the probability of events such as failure over a specified period. This is accomplished using the rate parameter \( \lambda \), linking back to the key property of the exponential distribution: the expected time until an event occurs (like failure) being constant, at the mean value given.

Survival Function

The survival function in the context of the exponential distribution helps us determine the probability of a device lasting longer than a specified time. The formula for the survival function is:

• \( P(T > t) = e^{-\lambda t} \)

This function tells us what the probability is that the device still operates after time \( t \), given the rate \( \lambda \). For example, calculating the probability that the device operates beyond three years:- First, plug \( \lambda = 0.2 \) and \( t = 3 \) into the survival function.- Calculate \( e^{-0.6} \), which approximately equals 0.5488.- Then \( P(T > 3) = 0.5488 \), meaning there's about a 54.88% chance the device will still be working after three years.By subtracting this survival probability from 1 (100%), we find the failure probability after three years, which is around 45.12%, offering insights into the expected reliability over time.

Memoryless Property

One remarkable characteristic of the exponential distribution is its memoryless property. This property implies that the probability of the device surviving for additional time, given it already survived a certain time, is the same as its original probability. Mathematically, this is denoted by:

• \( P(T > t+s | T > s) = P(T > t) \)

Where \( T \) is the time of event occurrence, \( t \) is any time period, and \( s \) is the time period that has already passed. In simple terms, past time does not affect future survival probability.For instance, if a device has been operational for 6 years, the probability it will function for another year (7 years total) is calculated the same way as if fresh. Using \( \lambda = 0.2 \) for one more year, the probability \( P(T > 1) = e^{-0.2} \approx 0.8187 \). Hence, the likelihood the device continues to work a seventh year remains approximately 81.87%, regardless of its six past operational years. This memoryless property simplifies many calculations in reliability engineering and queueing theory.