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The lifetime of a mechanical assembly in a vibration test is exponentially distributed with a mean of 400 hours. (a) What is the probability that an assembly on test fails in less than 100 hours? (b) What is the probability that an assembly operates for more than 500 hours before failure? (c) If an assembly has been on test for 400 hours without a failure, what is the probability of a failure in the next 100 hours?

Short Answer

Expert verified
(a) 0.2212; (b) 0.2865; (c) 0.2212.

Step by step solution

01

Understand the Exponential Distribution

The exponential distribution is used to model the time until an event occurs, like the failure of a mechanical assembly. It is characterized by the parameter \( \lambda \), which is the rate parameter. For a given mean \( \mu \), the rate parameter is \( \lambda = \frac{1}{\mu} \). Since the mean is 400 hours, we have \( \lambda = \frac{1}{400} = 0.0025 \).
02

Calculate Probability of Failure in Less Than 100 Hours

To find the probability that the assembly fails in less than 100 hours, we use the cumulative distribution function (CDF) for the exponential distribution: \( P(X < x) = 1 - e^{-\lambda x} \). Substituting \( \lambda = 0.0025 \) and \( x = 100 \), we get: \[ P(X < 100) = 1 - e^{-0.0025 \times 100} \approx 1 - e^{-0.25} \approx 1 - 0.7788 = 0.2212. \]
03

Calculate Probability of Operating More Than 500 Hours

To find the probability that the assembly operates for more than 500 hours, we use the complement rule: \( P(X > x) = 1 - P(X < x) \). First, calculate \( P(X < 500) \) using the CDF: \[ P(X < 500) = 1 - e^{-0.0025 \times 500} \approx 1 - e^{-1.25} \approx 1 - 0.2865 = 0.7135. \] Thus, \[ P(X > 500) = 1 - 0.7135 = 0.2865. \]
04

Conditional Probability After 400 Hours

Given that the assembly has survived 400 hours, we seek the probability of it failing in the next 100 hours. The memoryless property of the exponential distribution implies \( P(X > a + b \mid X > a) = P(X > b) \). Thus, \[ P(X > 500 \mid X > 400) = P(X > 100) = 1 - e^{-0.0025 \times 100} \approx 0.7788. \]The probability of failure in the next 100 hours is:\[ 1 - 0.7788 = 0.2212. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cumulative Distribution Function (CDF)
The Cumulative Distribution Function, or CDF, is a key concept in probability and statistics. It helps us determine the probability that a random variable, like the lifespan of a gadget or machine, takes on a value less than or equal to a specific point. For an exponential distribution, the CDF is given by the formula:\[ P(X < x) = 1 - e^{-\lambda x} \]- **\(P(X < x)\)** is the probability that the random variable \(X\) is less than \(x\).- **\(\lambda\)** is the rate parameter, telling you how quickly the event happens.- **\(e\)** is the base of the natural logarithm, approximately 2.718.In the context of our exercise, the CDF values help us figure out the chances of an assembly failing within a certain number of hours. For instance, computing the CDF for 100 hours using our rate parameter \(\lambda = 0.0025\), we get about 0.2212, meaning there's a 22.12% risk of failure in that time. Similarly, computing it for 500 hours gives us the probability of surviving or failing before that period.
Memoryless Property
The memoryless property is a unique feature of the exponential distribution that simplifies calculations in scenarios where time elapses without an event occurring. It implies that the future probability of an event occurring is the same now as it was initially, irrespective of past occurrences.Mathematically, the memoryless property states:- \(P(X > a + b \mid X > a) = P(X > b)\)This means if the assembly hasn't failed after 400 hours of testing, the probability it will last at least 100 more hours is the same as the probability of lasting 100 hours from the start, disregarding the time that has already passed.This property is useful in reliability testing, allowing engineers and technicians to predict failures over extended operational periods without recalculating the entire distribution. It underscores the idea that no history affects the probability of future failures.
Mean and Rate Parameter
Understanding the mean and the rate parameter is crucial in dealing with exponential distributions. The mean, represented as \(\mu\), indicates the expected average time until an event, like a failure, occurs. Here, it's given as 400 hours.The relationship between the mean and the rate parameter \(\lambda\) is captured in:\[ \lambda = \frac{1}{\mu} \]This formula means the rate parameter is the reciprocal of the mean. With a mean of 400 hours, the rate parameter becomes \(\lambda = 0.0025\). This parameter describes how densely packed or frequent the events (failures) are likely to occur.In practice, keeping an eye on these two metrics can help diagnose and plan for potential downtimes or servicing schedules in mechanical systems, based on statistical modeling. **By calculating using \(\lambda\), we can quickly solve related probabilities** like time to next failure or overall reliability over certain periods.

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