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Chapter 12: Problem 34

Recall that the median of an exponential density function is that number \(M\) such that \(\operatorname{Pr}(X \leq M)=\frac{1}{2} .\) Show that \(M=(\ln 2) / k .\) (We see that the median is less than the mean.)

Short Answer

Expert verified
The median is \(\frac{\ln 2}{k}\).

Step by step solution

01

Set up the Exponential Distribution Function

The probability density function (PDF) of an exponential distribution is given by: \[f(x) = ke^{-kx} \text{ for } x \geq 0\]We need to find the median, denoted as \(M\), such that the cumulative distribution function (CDF) at \(M\) equals to \(\frac{1}{2}\).
02

Define the Cumulative Distribution Function (CDF)

The CDF of an exponential distribution is: \[F(x) = 1 - e^{-kx}\text{ for } x \geq 0\]This function gives the probability that the random variable \(X\) takes on a value less than or equal to \(x\).
03

Set Up the Median Equation

By definition, the median \(M\) satisfies the equation: \[F(M) = \operatorname{Pr}(X\leq M) = \frac{1}{2}\]Therefore, we need to solve for \(M\) in the following equation: \[1 - e^{-kM} = \frac{1}{2}\]
04

Solve for M

Isolate the exponential term: \[1 - e^{-kM} = \frac{1}{2}\] Subtract 1 from both sides: \[- e^{-kM} = \frac{1}{2} - 1 = -\frac{1}{2}\] Multiply both sides by -1: \[e^{-kM} = \frac{1}{2}\] Apply the natural logarithm to both sides to solve for \(-kM\): \[-kM = \ln\left(\frac{1}{2}\right) = -\ln 2\]
05

Calculate the Median

Finally, solve for \(M\): \[-kM = -\ln 2\] Divide by \(-k\): \[M = \frac{\ln 2}{k}\] This shows that the median of the exponential distribution is \(\frac{\ln 2}{k}\).

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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) describes the likelihood of a random variable to take a particular value. For the exponential distribution, the PDF is given by:

\[ f(x) = ke^{-kx} \text{ for } x \geq 0 \]
Here, \( k \) is a constant that defines the rate of the distribution.

Key properties of a PDF include:

  • Non-negativity: \( f(x) \geq 0 \) for all \( x \).
  • Total area under the curve is 1: \( \int_0^{\infty} f(x) \, dx = 1 \).

The PDF helps us understand how data is distributed across different values. In the context of the exponential distribution, it tells us how likely it is that a given time until an event occurs, such as waiting time in a queue, falls within a certain range.

Cumulative Distribution Function
The cumulative distribution function (CDF) represents the probability that a random variable takes on a value less than or equal to a given value. For exponential distributions, the CDF is expressed as:

\[ F(x) = 1 - e^{-kx} \text{ for } x \geq 0 \]
This function helps us determine the likelihood that a particular event will occur within a certain timeframe.

Key features of the CDF include:

  • Monotonicity: It is a non-decreasing function.
  • Range: The CDF always lies between 0 and 1, inclusive.
  • Limits: As \( x \) goes to infinity, \( F(x) \) approaches 1; as \( x \) goes to zero, \( F(x) \) approaches 0.

In practice, the CDF is very useful for finding the median of a distribution, which is key to solving our original problem. We set the CDF equal to 0.5 (since the median splits the probability in half) and solve for the median \( M \). Thus, we solve:\

\[ 1 - e^{-kM} = \frac{1}{2} \].

Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \) where \( e \) is approximately 2.71828. It is widely used in mathematics, especially in growth processes like the exponential distribution.

Key properties of the natural logarithm include:

  • \( \ln(e) = 1 \).
  • \( \ln(1) = 0 \).
  • \( \ln(ab) = \ln(a) + \ln(b) \) for any positive \( a \) and \( b \).
  • \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
  • \( \ln(a^b) = b \ln(a) \).

In our solution, we use the natural logarithm to find \( M \). During the calculation, we take the logarithm of both sides of the equation \( e^{-kM} = \frac{1}{2} \) to solve for \( M \). Therefore,

\[ -kM = \ln\left(\frac{1}{2}\right) = -\ln(2) \].

Solving for \( M \) gives us:

\[ M = \frac{\ln 2}{k} \].

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Most popular questions from this chapter

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). $$f(x)=\frac{3 \sqrt{x}}{16}, 0 \leq x \leq 4$$

Time to Complete a Job When preparing a bid on a large construction project, a contractor analyzes how long each phase of the construction will take. Suppose that the contractor estimates that the time required for the electrical work will be \(X\) hundred worker-hours, where \(X\) is a random variable with density function \(f(x)=x(6-x) / 18,3 \leq x \leq 6.\) (See Fig. 3 ) (a) Find the cumulative distribution function \(F(x).\) (b) What is the likelihood that the electrical work will take less than 500 worker-hours? (c) Find the mean time to complete the electrical work. (d) Find \(\operatorname{Var}(X).\)

The number of times a printing press breaks down each month is Poisson distributed with \(\lambda=4 .\) What is the probability that the printing press breaks down between 2 and 8 times during a particular month?

The number of people arriving during a 5 -minute interval at a supermarket checkout counter is Poisson distributed with \(\lambda=8.\) (a) What is the probability that exactly eight people arrive during a particular 5 -minute period? (b) What is the probability that at most eight people arrive during a particular 5 -minute period?

The density function for a continuous random variable \(X\) on the interval \(1 \leq x \leq 4\) is \(f(x)=\frac{4}{9} x-\frac{1}{9} x^{2}.\) (a) Use \(f(x)\) to compute \(\operatorname{Pr}(3 \leq X \leq 4).\) (b) Find the corresponding cumulative distribution function \(F(x).\) (c) Use \(F(x)\) to compute \(\operatorname{Pr}(3 \leq X \leq 4).\)
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