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The probability density function of the time to failure of an electronic component in a copier (in hours) is \(f(x)=\) \(e^{-x / 1000 / 1000}\) for \(x>0 .\) Determine the probability that (a) A component lasts more than 3000 hours before failure. (b) A component fails in the interval from 1000 to 2000 hours. (c) A component fails before 1000 hours. (d) The number of hours at which \(10 \%\) of all components have failed

Short Answer

Expert verified
(a) 0.0498, (b) 0.1670, (c) 0.6321, (d) 105.36 hours.

Step by step solution

01

Understanding the Problem

We need to determine the probabilities related to the time to failure of an electronic component using the given probability density function (pdf). The pdf is a key to find the probability for the events defined in the question.
02

Identify the Probability Density Function

The pdf (probability density function) is given as \[ f(x) = \begin{cases} e^{-x / 1000 / 1000}, & x > 0 \ 0, & x \leq 0 \end{cases} \]This function represents an exponential distribution with a parameter \( \lambda = \frac{1}{1000} \). Here \( \lambda \) is the failure rate.
03

Calculate Probability for (a)

We want to find the probability that a component lasts more than 3000 hours. For an exponential distribution, the survival function is \[ P(X > x) = e^{-\lambda x} \]For \( x = 3000 \) hours, \[ P(X > 3000) = e^{-3000/1000} = e^{-3} \approx 0.0498 \]
04

Calculate Probability for (b)

For a failure within an interval \([1000, 2000]\) hours, we use the formula \[ P(1000 < X < 2000) = P(X < 2000) - P(X < 1000) \]Where \[ P(X < x) = 1 - e^{-\lambda x} \]Thus, \[ P(1000 < X < 2000) = (1 - e^{-2}) - (1 - e^{-1}) \approx 0.1353 - 0.6321 = 0.1670 \]
05

Calculate Probability for (c)

The probability that a component fails before 1000 hours is given by \[ P(X < 1000) = 1 - e^{-1000/1000} = 1 - e^{-1} \approx 0.6321 \]
06

Calculate Hours for 10% Fail Rate (d)

We need to find \( x \) such that \( P(X < x) = 0.10 \). This means solving \[ 1 - e^{-x/1000} = 0.10 \]Simplifying, \[ e^{-x/1000} = 0.90 \]Taking the natural logarithm on both sides, \[ -x/1000 = \ln(0.90) \] Thus, \[ x = -1000 \ln(0.90) \approx 105.36 \]
07

Conclusion

These steps lead to the following results: (a) Probability a component lasts more than 3000 hours is approximately 0.0498. (b) Probability it fails between 1000 and 2000 hours is approximately 0.1670. (c) Probability it fails before 1000 hours is approximately 0.6321. (d) 10% of all components will have failed by approximately 105.36 hours.

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Key Concepts

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

Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time until an event occurs, such as the failure of a component. This kind of distribution is characterized by its constant failure rate, known as the rate parameter \( \lambda \). The probability density function (pdf) for an exponential distribution is given by:
  • \( f(x) = \lambda e^{-\lambda x} \) for \( x > 0 \)
  • \( f(x) = 0 \) for \( x \leq 0 \)
In the context of our exercise, we have \( \lambda = \frac{1}{1000} \), meaning that 1 out of every 1000 units is expected to fail per hour on average.
This function is termed 'memoryless', implying that each additional hour a component works does not change the expected failure rate.
Survival Function
The survival function provides us with the probability that a component will survive past a certain time \( x \). It is fundamentally connected to the exponential distribution and is calculated using the following formula:
\[ P(X > x) = e^{-\lambda x} \]This function is particularly useful when determining the duration a component will last beyond specific milestones. In the exercise, to find out how likely a copier component can last beyond 3000 hours, we use the survival function yielding:
  • \( P(X > 3000) = e^{-3} \approx 0.0498 \)
This suggests there's a roughly 5% chance the component will operate past 3000 hours without failing.
Failure Rate
The failure rate, also known as the hazard rate, describes the instantaneous rate at which components fail. For the exponential distribution, this rate is constant and is simply the parameter \( \lambda \).
For our scenario, with \( \lambda = \frac{1}{1000} \), we determine that the component exhibits a roughly continuous and steady risk of failing per hour. This consistency is appealing in many fields such as reliability engineering because it simplifies calculations regarding expected lifespans and service requirements.
  • Constant failure rate: behavior indicative of the exponential distribution
  • Useful for planning maintenance or replacement schedules in organizations
Component Failure Probability
Understanding the probability of component failure at different times is vital in engineering and manufacturing. Calculating these probabilities allows for more strategic decision-making when it comes to warranty periods or predicting maintenance needs.
The task of determining failure probability between specific hours involves calculating the cumulative distribution function (CDF) differences:
  • For failures within an interval \([1000, 2000]\): \( P(1000 < X < 2000) \approx 0.1670 \)
  • Component failing before 1000 hours: \( P(X < 1000) \approx 0.6321 \)
These values guide how we might design product guarantees and anticipate customer satisfaction. Moreover, solving for the time when a specific percentage of components fail (in this case, 10%) helps in aligning the expectations with real-world durability, thus aiding in product lifecycle planning.

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Most popular questions from this chapter

Suppose that the length of stay (in hours) at a hospital emergency department is modeled with a lognormal random variable \(X\) with \(\theta=1.5\) and \(\omega=0.4\). Determine the following in parts (a) and (b): (a) Mean and variance (b) \(P(X<8)\) (c) Comment on the difference between the probability \(P(X<0)\) calculated from this lognormal distribution and a normal distribution with the same mean and variance.

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The diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch. (a) Suppose that the specifications require the dot diameter to be between 0.0014 and 0.0026 inch. If the probability that a dot meets specifications is to be \(0.9973,\) what standard deviation is needed? (b) Assume that the standard deviation of the size of a dot is 0.0004 inch. If the probability that a dot meets specifications is to be \(0.9973,\) what specifications are needed? Assume that the specifications are to be chosen symmetrically around the mean of 0.002 .

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