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Chapter 4: Problem 1

Suppose that \(X\) has an exponential distribution with mean equal to \(10 .\) Determine the following: a. \(P(X>10)\) b. \(P(X>20)\) c. \(P(X<30)\) d. Find the value of \(x\) such that \(P(X

Short Answer

Expert verified
a. 0.3679 b. 0.1353 c. 0.9502 d. 29.96

Step by step solution

01

Understanding the Exponential Distribution

An exponential distribution with mean \( \mu \) has the rate parameter \( \lambda = \frac{1}{\mu} \). Since the mean is 10, the rate parameter \( \lambda = \frac{1}{10} = 0.1 \). The probability density function for an exponential distribution is given as \( f(x) = \lambda e^{-\lambda x} \).
02

Calculate P(X>10)

The probability \( P(X>10) \) can be calculated using the complementary cumulative distribution function: \[ P(X>10) = 1 - P(X \le 10) = e^{-\lambda \(10\)} \]. Plug in \( \lambda = 0.1 \), we get: \[ P(X>10) = e^{-0.1 \cdot 10} = e^{-1} \approx 0.3679 \].
03

Calculate P(X>20)

Similarly, the probability \( P(X>20) \) is given by \[ P(X>20) = e^{-\lambda \(20\)} = e^{-0.1 \cdot 20} = e^{-2} \approx 0.1353 \].
04

Calculate P(X

The probability \( P(X<30) \) is found by using the cumulative distribution function: \[ P(X<30) = 1 - P(X>30) = 1 - e^{-\lambda \(30\)} \]. With \( \lambda = 0.1 \,\) we find: \[ P(X<30) = 1 - e^{-3} \approx 1 - 0.0498 = 0.9502 \].
05

Find the value of x such that P(X

To find \( x \) such that \( P(X<x) = 0.95 \,\) set the cumulative distribution to 0.95: \[ 1 - e^{-0.1x} = 0.95 \]. Solving for \( x \), we rearrange and simplify: \[ e^{-0.1x} = 0.05 \,\ \ -0.1x = \ln(0.05) \, \ x = \frac{\ln(0.05)}{-0.1} \approx 29.96 \].

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Key Concepts

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

Probability Density Function of the Exponential Distribution
The probability density function (pdf) is a crucial element of the exponential distribution. It describes the likelihood of a random variable taking on a specific value. In the context of the exponential distribution, the pdf is given by the formula:
  • \[ f(x) = \lambda e^{-\lambda x} \]
Here, \( \lambda \) is the rate parameter, which is calculated as the reciprocal of the mean \( \mu \). For instance, if the mean \( \mu = 10 \), then the rate \( \lambda = \frac{1}{10} = 0.1 \). This parameter signifies the frequency or rate at which events happen.
  • Interpretation: In a real-world context, this pdf helps in determining the probability of occurrences in a given period. For example, if you are measuring the time between phone calls at a call center, this function predicts how likely the next call will arrive in a particular amount of time.
  • Behavior: Since \( x \) represents time or distance in many applications, the function exponentially declines as \( x \) increases, meaning the likelihood of large intervals reduces in an exponential pattern.
Cumulative Distribution Function of the Exponential Distribution
The cumulative distribution function (cdf) helps determine the probability that a random variable is less than or equal to a certain value. For the exponential distribution, the cdf is represented as:
  • \[ F(x) = 1 - e^{-\lambda x} \]
This function accumulates the probabilities of all outcomes up to a particular value \( x \).
  • Usefulness: This is particularly useful when you want to calculate the probability of an event happening within a specific time frame. For example, to find the likelihood of receiving a product within 30 minutes, you would use the cdf to get \( P(X<30) \).
  • Example Calculation: Plugging in \( \lambda = 0.1 \) and \( x = 30 \) in the formula, you can determine that \( P(X<30) = 1 - e^{-3} \), which results in an approximate probability of 0.9502.
Complementary Cumulative Distribution Function of the Exponential Distribution
The complementary cumulative distribution function (ccdf), sometimes referred to as the survival function, calculates the probability that the random variable is greater than a particular value. This is simply the complement of the cdf:
  • \[ \text{CCDF}(x) = 1 - F(x) = e^{-\lambda x} \]
The ccdf is useful when you are interested in understanding the probability of an event not occurring within a given time frame.
  • Practicality: Suppose you want to determine the chance that a technician finishes a repair job in more than 20 minutes. You calculate \( P(X>20) \) using the ccdf formula \( e^{-2} \) and get an approximate result of 0.1353, meaning there's a 13.53% chance the job exceeds that time.
  • Complementary Nature: It's important to understand that since the cdf provides the probability of \( X \leq x \), the ccdf naturally focuses on the opposite scenario, capturing all possibilities where \( X > x \).

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Most popular questions from this chapter

An e-mail message will arrive at a time uniformly distributed between 9: 00 A.M. and 11: 00 A.M. You check e-mail at 9: 15 A.M. and every 30 minutes afterward. a. What is the standard deviation of arrival time (in minutes)? b. What is the probability that the message arrives less than 10 minutes before you view it? c. What is the probability that the message arrives more than 15 minutes before you view it?

Suppose that the cumulative distribution function of the random variable \(X\) is $$ F(x)=\left\\{\begin{array}{lr} 0 & x<0 \\ 0.25 x & 0 \leq x<5 \\ 1 & 5 \leq x \end{array}\right. $$ Determine the following: a. \(\quad P(X<2.8)\) b. \(P(X>1.5)\) c. \(P(X<-2)\) d. \(P(X>6)\)

The life of automobile voltage regulators has an exponential distribution with a mean life of 6 years. You purchase a 6-year-old automobile with a working voltage regulator and plan to own it for 6 years a. What is the probability that the voltage regulator fails during your ownership? b. If your regulator fails after you own the automobile 3 years and it is replaced, what is the mean time until the next failure?

An article in Microelectronics Reliability ["Advanced Electronic Prognostics through System Telemetry and Pattern Recognition Methods" \((2007,\) Vol. \(47(12),\) pp. \(1865-1873)]\) presented an example of electronic prognosis. The objective was to detect faults to decrease the system downtime and the number of unplanned repairs in high-reliability systems. Previous measurements of the power supply indicated that the signal is normally distributed with a mean of \(1.5 \mathrm{~V}\) and a standard deviation of \(0.02 \mathrm{~V}\) a. Suppose that lower and upper limits of the predetermined specifications are \(1.45 \mathrm{~V}\) and \(1.55 \mathrm{~V},\) respectively. What is the probability that a signal is within these specifications? b. What is the signal value that is exceeded with \(95 \%\) probability? c. What is the probability that a signal value exceeds the mean by two or more standard deviations?

The life of a semiconductor laser at a constant power is normally distributed with a mean of 7000 hours and a standard deviation of 600 hours. a. What is the probability that a laser fails before 5000 hours? b. What is the life in hours that \(95 \%\) of the lasers exceed? c. If three lasers are used in a product and they are assumed to fail independently, what is the probability that all three are still operating after 7000 hours?
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