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Chapter 6: Problem 54

The time (in minutes) between telephone calls at an insurance claims office has the following exponential probability distribution. \\[ f(x)=.50 e^{-.50 x} \quad \text { for } x \geq 0 \\] a. What is the mean time between telephone calls? b. What is the probability of having 30 seconds or less between telephone calls? c. What is the probability of having 1 minute or less between telephone calls? d. What is the probability of having 5 or more minutes without a telephone call?

Short Answer

Expert verified
a. Mean is 2 minutes. b. Probability for 30 seconds or less is 0.2212. c. Probability for 1 minute or less is 0.3935. d. Probability for 5 or more minutes is 0.0821.

Step by step solution

01

Identify Mean of Exponential Distribution

In an exponential distribution, the mean is given by \( \frac{1}{\lambda} \), where \( \lambda \) is the rate parameter. In this problem, \( \lambda = 0.50 \). Therefore, the mean time between calls is \( \frac{1}{0.50} = 2 \) minutes.
02

Calculate Probability for 30 Seconds or Less

Probability is determined by the cumulative distribution function (CDF) for the exponential distribution, which is \( F(x) = 1 - e^{-\lambda x} \). For 30 seconds, we use \( x = 0.5 \) minutes (since 30 seconds is half a minute). Therefore, \[ F(0.5) = 1 - e^{-0.50 \times 0.5} = 1 - e^{-0.25} \approx 0.2212. \] Hence, the probability is approximately 0.2212.
03

Calculate Probability for 1 Minute or Less

Using the same CDF for \( x = 1 \) minute, \[ F(1) = 1 - e^{-0.50 \times 1} = 1 - e^{-0.50} \approx 0.3935. \] Thus, the probability of having 1 minute or less between calls is approximately 0.3935.
04

Calculate Probability for 5 or More Minutes

To find the probability of 5 or more minutes without a call, we use the complement of the CDF up to 5 minutes, \[ P(X \geq 5) = 1 - F(5) = e^{-0.50 \times 5} = e^{-2.5} \approx 0.0821. \] Therefore, the probability is approximately 0.0821.

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Key Concepts

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

Probability Distribution
An exponential probability distribution is useful for modeling situations where events occur continuously and independently. This type of distribution is often used to model the time between events in a process that occurs at a constant average rate.
It is a continuous distribution meaning it deals with variables that can take any value within a specified range.

Key Characteristics:
  • Describes the time or space between events in a Poisson process.
  • The probability density function (PDF) for the exponential distribution is given by: \( f(x) = \lambda e^{-\lambda x} \) for \( x \ge 0 \), where \( \lambda \) is the rate parameter.
  • The exponential distribution is often memoryless, meaning the probability of an event occurring in the future is independent of any past events.
Understanding how exponential distributions work can help in numerous fields like finance, insurance, and telecommunications.
Cumulative Distribution Function
The cumulative distribution function (CDF) for an exponential distribution gives the probability that the random variable \( X \) is less than or equal to a particular value. It helps in quantifying the accumulated probability up to a specific point.
For an exponential distribution, the CDF is given by: \[ F(x) = 1 - e^{-\lambda x} \]

Essential Points:
  • The CDF measures the total probability accumulated from the left end of the distribution up to a specific value \( x \).
  • This function is particularly valuable for calculating probabilities for ranges rather than exact values.
  • Functions like \( F(0.5) = 1 - e^{-0.25} \) help estimate probabilities for real-world scenarios, like the time between phone calls.
By using the CDF, one can efficiently calculate probabilities of events within a given interval in an exponential setting.
Rate Parameter
The rate parameter, denoted as \( \lambda \), is a critical part of the exponential distribution. It describes how frequently events occur.
A larger \( \lambda \) indicates more frequent events, while a smaller \( \lambda \) suggests infrequent events.

Important Considerations:
  • In the given example, \( \lambda = 0.50 \), indicating relatively infrequent calls are expected.
  • The rate parameter is the reciprocal of the mean (average duration between events): \( \lambda = \frac{1}{\text{mean}} \).
  • The \( \lambda \) parameter directly affects both the probability density function (PDF) and the cumulative distribution function (CDF) shapes.
Understanding the rate parameter can aid in effectively interpreting the behavior of the exponential distribution in real-world applications.
Mean Calculation
In exponential distributions, calculating the mean is straightforward and highly useful.
The mean of an exponential distribution indicates the average time or space between occurrences of the event.
For exponential distributions, the mean is the reciprocal of the rate parameter \( \lambda \):\[ \text{Mean} = \frac{1}{\lambda} \]

Real-World Applications:
  • In the exercise, \( \lambda = 0.50 \), so the mean time between telephone calls is \( \frac{1}{0.50} = 2 \) minutes.
  • A mean calculation helps understand the typical wait time or distance between events in processes like arrivals of calls, machine failures, etc.
  • Understanding the mean can assist in resource allocation and preparing for variable demands.
The mean is a powerful metric for planning and analysis, allowing for well-informed logistical and operational decisions.

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Most popular questions from this chapter

A business executive, transferred from Chicago to Atlanta, needs to sell her house in Chicago quickly. The executive's employer has offered to buy the house for \(\$ 210,000\) but the offer expires at the end of the week. The executive does not currently have a better offer but can afford to leave the house on the market for another month. From conversations with her realtor, the executive believes the price she will get by leaving the house on the market for another month is uniformly distributed between \(\$ 200,000\) and \(\$ 225,000\) a. If she leaves the house on the market for another month, what is the mathematical expression for the probability density function of the sales price? b. If she leaves it on the market for another month, what is the probability that she will get at least \(\$ 215,000\) for the house? c. If she leaves it on the market for another month, what is the probability that she will get less than \(\$ 210,000 ?\) d. Should the executive leave the house on the market for another month? Why or why not?

Comcast Corporation is the largest cable television company, the second largest Internet service provider, and the fourth largest telephone service provider in the United States. Generally known for quality and reliable service, the company periodically experiences unexpected service interruptions. On January \(14,2009,\) such an interruption occurred for the Comcast customers living in southwest Florida. When customers called the Comcast office, a recorded message told them that the company was aware of the service outage and that it was anticipated that service would be restored in two hours. Assume that two hours is the mean time to do the repair and that the repair time has an exponential probability distribution. a. What is the probability that the cable service will be repaired in one hour or less? b. What is the probability that the repair will take between one hour and two hours? c. For a customer who calls the Comcast office at 1: 00 P.M., what is the probability that the cable service will not be repaired by 5: 00 p.M.?

The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. What is the probability that the arrival time between vehicles is 12 seconds or less? c. What is the probability that the arrival time between vehicles is 6 seconds or less? d. What is the probability of 30 or more seconds between vehicle arrivals?

An Internal Revenue Oversight Board survey found that \(82 \%\) of taxpayers said that it was very important for the Internal Revenue Service (IRS) to ensure that high-income taxpayers do not cheat on their tax returns (The Wall Street Journal, February 11,2009 ). a. For a sample of eight taxpayers, what is the probability that at least six taxpayers say that it is very important to ensure that high-income taxpayers do not cheat on their tax returns? Use the binomial distribution probability function shown in Section 5.4 to answer this question. b. For a sample of 80 taxpayers, what is the probability that at least 60 taxpayers say that it is very important to ensure that high-income taxpayers do not cheat on their tax returns? Use the normal approximation of the binomial distribution to answer this question. c. As the number of trails in a binomial distribution application becomes large, what is the advantage of using the normal approximation of the binomial distribution to compute probabilities? d. When the number of trials for a binominal distribution application becomes large, would developers of statistical software packages prefer to use the binomial distribution probability function shown in Section 5.4 or the normal approximation of the binomial distribution shown in Section 6.3? Explain.

A random variable is normally distributed with a mean of \(\mu=50\) and a standard deviation of \(\sigma=5\) a. Sketch a normal curve for the probability density function. Label the horizontal axis with values of \(35,40,45,50,55,60,\) and \(65 .\) Figure 6.4 shows that the normal curve almost touches the horizontal axis at three standard deviations below and at three standard deviations above the mean (in this case at 35 and 65 ). b. What is the probability that the random variable will assume a value between 45 and \(55 ?\) c. What is the probability that the random variable will assume a value between 40 and \(60 ?\)
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