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Chapter 4: Problem 129

If the random variable \(X\) has an exponential distribution with mean \(\theta,\) determine the following: (a) \(P(X>\theta)\) (b) \(P(X>2 \theta)\) (c) \(P(X>3 \theta)\) (d) How do the results depend on \(\theta\) ?

Short Answer

Expert verified
(a) \(e^{-1}\), (b) \(e^{-2}\), (c) \(e^{-3}\), (d) Probabilities are determined by \( a/\theta \) ratio, not \( \theta \) itself.

Step by step solution

01

Understanding the Exponential Distribution

An exponential distribution with mean \( \theta \) has a probability density function (PDF) given by \( f(x;\theta) = \frac{1}{\theta} e^{-x/\theta} \) for \( x \geq 0 \). Here, \( \theta \) is the parameter of the distribution, which is also the mean. A key property of the exponential distribution is that it is 'memoryless', meaning that \( P(X > a + b \mid X > a) = P(X > b) \). We can use the cumulative distribution function (CDF) to calculate the probabilities.
02

Deriving the CDF for the Exponential Distribution

The cumulative distribution function, \( F(x; \theta) \), is \( F(x; \theta) = 1 - e^{-x/\theta} \). We use this to find the probability \( P(X > a) = 1 - F(a; \theta) = e^{-a/\theta}\).
03

Calculate \( P(X > \theta) \)

Substitute \( a = \theta \) into the formula for \( P(X > a) \):\[ P(X > \theta) = e^{-\theta/\theta} = e^{-1}. \]
04

Calculate \( P(X > 2\theta) \)

Substitute \( a = 2\theta \) into the formula: \[ P(X > 2\theta) = e^{-2\theta/\theta} = e^{-2}. \]
05

Calculate \( P(X > 3\theta) \)

Substitute \( a = 3\theta \) into the formula: \[ P(X > 3\theta) = e^{-3\theta/\theta} = e^{-3}. \]
06

Examine the Dependence on \( \theta \)

The probabilities \( P(X > a) = e^{-a/\theta} \) are expressed in terms of the ratio \( a/\theta \). Thus, changing \( \theta \) scales the \( X \)-axis of the probability distribution, but the formula depends only on the ratio \( a/\theta \). Therefore, the specific values of the probabilities are independent of \( \theta \) as long as \( a \) is a multiple of \( \theta \).

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Key Concepts

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

Probability Density Function
The probability density function (PDF) of the exponential distribution provides the probability of the continuous random variable, which in this case is represented by the time until an event occurs. For an exponential distribution with a mean \( \theta \), the PDF is given by the formula:
  • \( f(x;\theta) = \frac{1}{\theta} e^{-x/\theta} \) for \( x \geq 0 \).
  • This tells us the likelihood of the random variable \( X \) taking on a precise value.
  • It is mapped over non-negative real numbers \( x \). In other words, the exponential distribution is defined for \( x \geq 0 \).
The parameter \( \theta \) controls the rate at which events occur, also known as the 'rate parameter.' This PDF is fundamental as it helps in understanding the rate of occurrences per time unit. The shape of this distribution has a constant decay rate, characterized by its memoryless property.
Cumulative Distribution Function
The cumulative distribution function (CDF) of an exponential distribution helps us determine the probability that a random variable \( X \) is less than or equal to a certain value. For the exponential distribution, this is computed as:
  • \( F(x; \theta) = 1 - e^{-x/\theta} \).
  • This expression gives us a cumulative probability, building up as \( x \) increases, starting at zero and approaching one.
Through the CDF, determining probabilities such as \( P(X > a) \) is straightforward:- The idea is to calculate \( 1 - F(a; \theta) = e^{-a/\theta}\). This allows us to handle situations where we need to know the chance of a random variable being greater than a certain point.
Mean of Exponential Distribution
The mean of the exponential distribution is simply the parameter \( \theta \). This mean dictates the average time or space before an occurrence happens within the context of the distribution:
  • It is a central measure of the distribution, acting both as the expected value of \( X \) and inverse of the rate at which events happen.
  • Understanding and calculating the mean helps us understand how spread the distribution of times is.
  • The mean \( \theta \) directly influences the spread of the distribution. A larger \( \theta \) implies a slower rate of events, making the event occurrence less frequent.
In practice, being able to calculate \( \theta \) allows for practical applications in areas like customer service times, failure rates, and other time-to-occurrence scenarios.
Memoryless Property
The memoryless property is a unique characteristic of the exponential distribution, meaning the future probability of an event does not depend on any past events:
  • Formally, this is expressed as \( P(X > a + b \mid X > a) = P(X > b) \).
  • This indicates that, regardless of how much time has already elapsed, the probability of observing an occurrence beyond a given point does not change.
  • Practically, this makes the exponential distribution ideal for modeling situations where "memory" of past events doesn't influence future ones, such as the time until the next customer service call ends, given it's ongoing.
This property, while seeming counterintuitive, is practical and simplifies calculations within various statistical modeling scenarios.

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Most popular questions from this chapter

Assume that the life of a roller bearing follows a Weibull distribution with parameters \(\beta=2\) and \(\delta=10,000\) hours. (a) Determine the probability that a bearing lasts at least 8000 hours. (b) Determine the mean time until failure of a bearing. (c) If 10 bearings are in use and failures occur independently, what is the probability that all 10 bearings last at least 8000 hours?

An airline makes 200 reservations for a flight that holds 185 passengers. The probability that a passenger arrives for the flight is \(0.9,\) and the passengers are assumed to be independent. (a) Approximate the probability that all the passengers who arrive can be seated. (b) Approximate the probability that the flight has empty seats. (c) Approximate the number of reservations that the airline should allow so that the probability that everyone who arrives can be seated is \(0.95 .\) [Hint: Successively try values for the number of reservations.]

An article in Vaccine ["Modeling the Effects of Influenza Vaccination of Health Care Workers in Hospital Departments" (2009, Vol.27(44), pp. \(6261-6267\) ) ] considered the immunization of healthcare workers to reduce the hazard rate of influenza virus infection for patients in regular hospital departments. In this analysis, each patient's length of stay in the department is taken as exponentially distributed with a mean of 7.0 days. (a) What is the probability that a patient stays in hospital for less than 5.5 days? (b) What is the probability that a patient stays in hospital for more than 10.0 days if the patient has currently stayed for 7.0 days? (c) Determine the mean length of stay such that the probability is 0.9 that a patient stays in the hospital less than 6.0 days.

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? (d) If 10 assemblies are tested, what is the probability that at least one fails in less than 100 hours? Assume that the assemblies fail independently. (e) If 10 assemblies are tested, what is the probability that all have failed by 800 hours? Assume that the assemblies fail independently.

Assume that the life of a packaged magnetic disk exposed to corrosive gases has a Weibull distribution with \(\beta=0.5\) and the mean life is 600 hours. Determine the following: (a) Probability that a disk lasts at least 500 hours. (b) Probability that a disk fails before 400 hours.
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