/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 114 Suppose that \(X\) has an expone... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(X\) has an exponential distribution with a mean of \(10 .\) Determine the following: (a) \(P(X<5)\) (b) \(P(X<15 \mid X>10)\) (c) Compare the results in parts (a) and (b) and comment on the role of the lack of memory property.

Short Answer

Expert verified
(a) 0.3935; (b) 0.3935; Lack of memory property shows P(X<15 | X>10) is same as P(X<5).

Step by step solution

01

Understand the Exponential Distribution

The exponential distribution is often used to model waiting times and is defined by its rate parameter \( \lambda \). For an exponential distribution, the mean is given by \( \frac{1}{\lambda} \). Given that the mean is \( 10 \), we find the rate parameter: \( \lambda = \frac{1}{10} = 0.1 \).
02

Solve for (a): Compute \(P(X

The cumulative distribution function (CDF) for an exponential distribution is given by \( 1 - e^{-\lambda x} \). Substituting \( \lambda = 0.1 \) and \( x = 5 \), we find \( P(X<5) = 1 - e^{-0.1 \times 5} = 1 - e^{-0.5} \). Evaluate this using a calculator to find \( 1 - 0.6065 = 0.3935 \).
03

Solve for (b): Compute \(P(X10)\)

The lack of memory property of the exponential distribution states that \( P(X < x + t \mid X > x) = P(X < t) \). Hence, \( P(X<15 \mid X>10) = P(X<5) \). So the probability is the same as in part (a), which is \( 0.3935 \).
04

Compare (a) and (b) - Comment on Lack of Memory

Comparing parts (a) and (b), we see that \( P(X<5) \) and \( P(X<15 \mid X>10) \) yield the same result. This illustrates the lack of memory property, which shows that the future probability \( P(X<15 \mid X>10) \) is independent of the past value and is solely based on the remaining interval \( x \) after surpassing \( 10 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Lack of Memory Property
The lack of memory property is a fascinating aspect of the exponential distribution. It basically means that the probability of an event occurring in the future is independent of the past events. No matter what has happened before, the exponential distribution "forgets" about it and focuses only on what's to come. This is why it's used to model processes that are random but consistent over time.

In mathematical terms, this property can be expressed as:
  • \( P(X < x + t \mid X > x) = P(X < t) \)
This equation means that the probability of experiencing an event in the next \( t \) units of time is the same no matter how long you've already been waiting. This was illustrated in our exercise by showing that \( P(X<15 \mid X>10) = P(X<5) \). The memoryless nature of the exponential distribution is precisely what makes it unique compared to other probability distributions.

So, when you're waiting for something that is exponentially distributed, it doesn't matter how long you've already waited. The future remains unaffected by the past.
Cumulative Distribution Function
The cumulative distribution function (CDF) is crucial for understanding the behavior of random variables. In the context of the exponential distribution, the CDF helps in determining the probability of a random variable being less than a certain value.

For an exponential distribution with rate parameter \( \lambda \), the CDF is given by:
  • \( F(x) = 1 - e^{-\lambda x} \)
This equation provides the probability that the random variable takes on a value less than or equal to \( x \). It is the area under the probability density function from 0 to \( x \).

In our problem, we used the CDF to find \( P(X<5) \). By substituting \( \lambda = 0.1 \) and \( x = 5 \) into the formula, we found that \( P(X<5) = 0.3935 \). The CDF is a powerful tool for assessing probabilities and understanding how a random variable is distributed.
Exponential Distribution Rate Parameter
The rate parameter, denoted by \( \lambda \), is a key part of the exponential distribution. It determines how "fast" or "slow" events are expected to occur. A larger \( \lambda \) indicates a higher rate of occurrence, meaning events happen more frequently.

For an exponential distribution, the mean, \( 1/\lambda \), is a crucial connection between \( \lambda \) and expected waiting time. In our exercise, the mean was provided as 10, leading to a \( \lambda \) of 0.1:
  • \( \lambda = \frac{1}{10} = 0.1 \)
This means that on average, we'd expect an event to occur every 10 units of time. Understanding \( \lambda \) helps in contextualizing the likelihood of an event happening within a certain timeframe.

The rate parameter is a vital concept because it offers insight into the frequency and predictability of events in the exponential distribution, making it useful in fields such as reliability engineering and queueing theory.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the regional right ventricle transverse wall motion in patients with pulmonary hypertension (PH). The right ventricle ejection fraction (EF) is approximately normally distributed with standard deviation of 12 for \(\mathrm{PH}\) subjects, and with mean and standard deviation of 56 and \(8,\) respectively, for control subjects. (a) What is the EF for control subjects exceeded with \(99 \%\) probability? (b) What is the mean for PH subjects such that the probability is \(1 \%\) that the EF of a PH subject is greater than the value in part (a)? (c) Comment on how well the control and PH subjects [with the mean determined in part (b)] can be distinguished by EF measurements.

Without an automated irrigation system, the height of plants two weeks after germination is normally distributed with a mean of 2.5 centimeters and a standard deviation of 0.5 centimeter. (a) What is the probability that a plant's height is greater than 2.25 centimeters? (b) What is the probability that a plant's height is between 2.0 and 3.0 centimeters?

The life of automobile voltage regulators has an exponential distribution with a mean life of six years. You purchase a six-year-old automobile, with a working voltage regulator and plan to own it for six years. (a) What is the probability that the voltage regulator fails during your ownership? (b) If your regulator fails after you own the automobile three years and it is replaced, what is the mean time until the next failure?

An article in Ad Hoc Networks ["Underwater Acoustic Sensor Networks: Target Size Detection and Performance Analysis" \((2009,\) Vol. \(7(4),\) pp. \(803-808)]\) discussed an underwater acoustic sensor network to monitor a given area in an ocean. The network does not use cables and does not interfere with shipping activities. The arrival of clusters of signals generated by the same pulse is taken as a Poisson arrival process with a mean of \(\lambda\) per unit time. Suppose that for a specific underwater acoustic sensor network, this Poisson process has a rate of 2.5 arrivals per unit time. (a) What is the mean time between 2.0 consecutive arrivals? (b) What is the probability that there are no arrivals within 0.3 time units? (c) What is the probability that the time until the first arrival exceeds 1.0 unit of time? (d) Determine the mean arrival rate such that the probability is 0.9 that there are no arrivals in 0.3 time units.

Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the mean and variance of \(X\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.