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After measuring the duration of many telephone calls, the telephone company found their data was well approximated by the density function \(p(x)=0.4 e^{-0.4 x}\) where \(x\) is the duration of a call, in minutes. (a) What percentage of calls last between 1 and 2 minutes? (b) What percentage of calls last 1 minute or less? (c) What percentage of calls last 3 minutes or more? (d) Find the cumulative distribution function.

Short Answer

Expert verified
(a) 24.5%; (b) 33%; (c) 30.1%; (d) \(F(x) = 1 - e^{-0.4x}\).

Step by step solution

01

Understand the Problem

We're given a probability density function (PDF) for call durations: \(p(x) = 0.4 e^{-0.4 x}\). We need to calculate the probabilities for various call duration ranges and find the cumulative distribution function (CDF).
02

Define the Cumulative Distribution Function (CDF)

The cumulative distribution function \(F(x)\) is defined as \(F(x) = \int_{0}^{x} p(t) \, dt\). This function gives us the probability that a call lasts less than or equal to \(x\) minutes.
03

Integrate to find the CDF

Let's find the CDF by integrating the PDF:\[ F(x) = \int_{0}^{x} 0.4 e^{-0.4 t} \, dt \]The integral of the exponential function is:\[ F(x) = -e^{-0.4 t} \bigg|_{0}^{x} \]Calculating the definite integral, we have:\[ F(x) = -e^{-0.4x} + e^0 = 1 - e^{-0.4x} \]
04

Find Probability for (a) 1 to 2 minutes

Use the CDF to find \( P(1 < x < 2) \):\[ P(1 < x < 2) = F(2) - F(1) \]\[ F(2) = 1 - e^{-0.8}, \quad F(1) = 1 - e^{-0.4} \]\[ P(1 < x < 2) = (1 - e^{-0.8}) - (1 - e^{-0.4}) = e^{-0.4} - e^{-0.8} \]
05

Calculate Percentage for (a)

Calculate the probability and convert it to a percentage:\[ P(1 < x < 2) = e^{-0.4} - e^{-0.8} \approx 0.245 \]So, the percentage is approximately 24.5%.
06

Find Probability for (b) 1 minute or less

Use the CDF to find \( P(x \leq 1) \):\[ P(x \leq 1) = F(1) = 1 - e^{-0.4} \]Calculating, we have:\[ P(x \leq 1) = 1 - e^{-0.4} \approx 0.33 \]
07

Calculate Percentage for (b)

Convert the probability to a percentage:\[ P(x \leq 1) \approx 0.33 \times 100\% = 33\% \]
08

Find Probability for (c) 3 minutes or more

Use the complementary probability to find \( P(x \geq 3) \) using the CDF:\[ P(x \geq 3) = 1 - F(3) = 1 - (1 - e^{-1.2}) = e^{-1.2} \]
09

Calculate Percentage for (c)

Convert the probability to a percentage:\[ P(x \geq 3) = e^{-1.2} \approx 0.301 \times 100\% \approx 30.1\% \]
10

Final CDF Expression (d)

We have already determined the CDF:\[ F(x) = 1 - e^{-0.4x} \] This provides the cumulative distribution of the call durations.

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Key Concepts

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

Cumulative Distribution Function Explained
The cumulative distribution function (CDF) is a crucial concept in probability and statistics. It represents the probability that a random variable will take a value less than or equal to a specific value. In other words, for a duration of call, the CDF tells us the likelihood that the call lasts no longer than that amount of time.
The formula for the CDF, based on a given probability density function (PDF), is written as:
  • \( F(x) = \int_{0}^{x} p(t) \, dt \)
This integral accumulates the probabilities from zero to \(x\), forming a function that starts at zero and increases to one as \(x\) increases. For the exercise given, the PDF is \( p(x) = 0.4 e^{-0.4 x} \), leading to the CDF:
  • \( F(x) = 1 - e^{-0.4x} \).
This CDF indicates how likely it is for a call's duration to be less than or equal to any given time \(x\). This step is fundamental to solve further probability questions by providing cumulative probabilities over a range of times.
Understanding Exponential Distribution
The exponential distribution is a probability distribution that describes time between events in a Poisson process, where events occur continuously and independently at a constant average rate. It's often used to model the time until an event happens, such as the duration of a phone call until it ends.

In the case of this exercise, the duration of phone calls follows an exponential distribution described by the density function \( p(x) = 0.4 e^{-0.4 x} \). This type of distribution is characterized by its memoryless property, meaning that the likelihood of a future event occurring does not depend on past behavior.

Key characteristics of exponential distribution include:
  • The mean, which is the average time between events. For the given PDF, this is \( \frac{1}{0.4} = 2.5 \) minutes.
  • The rate parameter, often denoted by \( \lambda \), which in this context is \(0.4\) per minute.
Understanding these characteristics of the exponential distribution helps us comprehend various scenarios' probabilities and deal with problems like calculating how long a phone call might take.
Probability Calculation Made Simple
Probability calculation involves determining the chance that a certain event will occur, often expressed as a percentage. To solve problems using the exponential distribution, like those in this exercise, you can utilize the cumulative distribution function (CDF).

Steps involve:
  • Identifying the range of interest, such as calls lasting between 1 and 2 minutes.
  • Calculating \( F(x) \) at the range endpoints using the CDF formula. For example, \( F(2) - F(1) \) provides the probability of calls between 1 to 2 minutes.
  • For a range starting from zero, the end probability is simply \( F(x) \).
  • To find probabilities for intervals starting above zero, like 3 minutes and more, utilize complementary probability: \( 1 - F(x) \).
For the provided solutions:
  • Cumulative probability \(P(x \leq 1)\) gives a 33% chance for calls lasting 1 minute or less.
  • Probability of calls lasting between 1 and 2 minutes was approximately 24.5%.
  • Calculating for calls lasting 3 or more minutes yields a 30.1% probability.
By systematically using the rules of probability calculation, students can effectively analyze and solve a range of similar exercises across various applications.

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