Suppose that a number of miles that a car can run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5,000-mile trip, what is the probability that she will be able to complete her trip without having to replace her car battery? What can be said when the distribution is not exponential?
The Correct Answer and Explanation is:
Let’s analyze the problem step by step.
Problem Recap:
- The lifetime of a car battery (in miles) is exponentially distributed.
- The average lifetime is 10,000 miles.
- The person wants to take a 5,000-mile trip.
- We want the probability that the battery lasts at least 5,000 miles.
Step 1: Identify the distribution parameters
The exponential distribution is defined by the probability density function (PDF): f(x)=λe−λxfor x≥0f(x) = \lambda e^{-\lambda x} \quad \text{for } x \ge 0
where λ\lambda is the rate parameter.
The mean of the exponential distribution is: E[X]=1λE[X] = \frac{1}{\lambda}
Given the average lifetime E[X]=10,000E[X] = 10,000, we have: λ=110,000=0.0001\lambda = \frac{1}{10,000} = 0.0001
Step 2: Find the probability battery lasts at least 5,000 miles
The exponential distribution’s survival function (probability that X>xX > x) is: P(X>x)=e−λxP(X > x) = e^{-\lambda x}
For x=5,000x = 5,000: P(X>5000)=e−0.0001×5000=e−0.5≈0.6065P(X > 5000) = e^{-0.0001 \times 5000} = e^{-0.5} \approx 0.6065
Answer:
The probability that the car battery will last for the entire 5,000-mile trip is approximately 0.6065, or 60.65%.
Step 3: What if the distribution is not exponential?
The exponential distribution is special because it has the memoryless property: P(X>s+t∣X>s)=P(X>t)P(X > s + t \mid X > s) = P(X > t)
This means that the probability of lasting another tt miles does not depend on how many miles have already been driven. The lifetime “resets” at any point, simplifying reliability analysis.
If the battery lifetime is not exponentially distributed, then:
- The memoryless property does not hold.
- The probability of lasting 5,000 miles could depend on how long the battery has already lasted.
- The survival function P(X>x)P(X > x) would come from the actual distribution, which might be more complicated (e.g., Weibull, normal, lognormal).
- Batteries often wear out more predictably as they age (e.g., increasing failure rate over time), so non-exponential distributions might better model aging or wear-out phenomena.
- In those cases, reliability calculations become more complex because past usage affects future survival probability.
- Without the memoryless property, one cannot simply use P(X>x)=e−λxP(X > x) = e^{-\lambda x} but must know the exact distribution and its parameters.
Summary Explanation:
In this problem, the exponential distribution assumption means the car battery’s lifetime is random with a constant failure rate and the memoryless property. Since the mean lifetime is 10,000 miles, the probability that the battery lasts for a 5,000-mile trip is about 60.65%. This uses the survival function of the exponential distribution, P(X>5000)=e−0.5P(X > 5000) = e^{-0.5}.
The exponential distribution is often a first model for lifetimes because it simplifies reliability calculations due to its memoryless property. However, real batteries might degrade over time, meaning their failure rate changes (often increases). This implies a non-exponential distribution might be more accurate, such as Weibull, which models increasing failure rates and aging. In such cases, the probability of lasting 5,000 miles depends on the battery’s current age or miles used and requires the full distribution’s survival function to evaluate. Without the memoryless property, the problem is more complex, and the simple formula e−λxe^{-\lambda x} is no longer valid.
If you’d like, I can also explain how to calculate survival probabilities for other distributions!