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Failure time The time between failures of a photocopier is exponentially distributed. Half of the copiers at a university require service during the first 2 years of operations. If the university purchased 150 copiers, how many do you expect to require service during the first year of their operation?

Short Answer

Expert verified
About 44 copiers are expected to require service during the first year.

Step by step solution

01

Understand the Exponential Distribution

The problem states that the time between failures of a photocopier follows an exponential distribution. This means we have a random variable representing time to failure with the probability density function \( f(t) = \lambda e^{-\lambda t} \), where \( \lambda \) is the rate parameter.
02

Determine the Median Using Given Data

We know that half of the copiers require service within 2 years, which means the median of the distribution is 2 years. The median for an exponential distribution is given by \( \ln(2)/\lambda \). Set this equal to 2 to find \( \lambda \): \[ \ln(2)/\lambda = 2 \]Thus, \( \lambda = \ln(2)/2 \).
03

Calculate the Probability of Failure in One Year

Using the parameter \( \lambda \), the probability that a copier fails within one year is given by the cumulative distribution function (CDF) of the exponential distribution: \[ P(T < 1) = 1 - e^{-\lambda \times 1} \]Substitute \( \lambda = \ln(2)/2 \) into this equation: \[ P(T < 1) = 1 - e^{-\ln(2)/2} \]
04

Simplify the Exponential Expression

By simplifying the expression, we get: \[ P(T < 1) = 1 - e^{-\ln(2)/2} \]This simplifies to: \[ P(T < 1) = 1 - (e^{\ln(2)})^{-1/2} = 1 - \sqrt{0.5} \]Calculate further: \[ P(T < 1) = 1 - \frac{1}{\sqrt{2}} \approx 0.2929 \]
05

Calculate Expected Number of Copiers Needing Service

Now, calculate the expected number of copiers needing service in the first year by multiplying the total number of copiers by the probability calculated: \[ \text{Expected Copiers} = 150 \times 0.2929 \approx 43.94 \]Approximately 44 copiers are expected to require service during the first year.

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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) is a fundamental concept used to describe the likelihood of a continuous random variable. For an exponential distribution, the PDF is expressed as \( f(t) = \lambda e^{-\lambda t} \), where \( \lambda \) is a rate parameter indicating the frequency or rate of events occurring. In this case, \( t \) represents time, such as the time between failures of a photocopier.
The exponential distribution is characterized by being "memoryless," meaning the probability of an event occurring in the future is independent of past events. This is particularly useful in scenarios where failures or arrivals occur continuously and independently over time.
Using the PDF helps determine the likelihood of failure or any event happening at a precise time. This contrasts with the cumulative distribution function (CDF), which provides the probability of the event occurring at any time until \( t \). Both functions together give a complete picture of the distribution of time-to-event variables.
Understanding the PDF allows you to model real-world processes, such as calculating the chance for machines breaking down in a specified timeframe. This is key to planning maintenance and resource allocation efficiently.
Median of Exponential Distribution
The median is a statistical measure representing the midpoint of a data set. In the context of an exponential distribution, it is the time by which 50% of the events (e.g., failures) will have occurred.
The formula for the median in an exponential distribution is given by \( \ln(2)/\lambda \), where \( \lambda \) is the rate parameter of the distribution. For the exercise provided, the median is specified as 2 years. Therefore, we can say that by the end of year two, half of the photocopiers would require servicing.
Using the formula, \( \ln(2)/\lambda = 2 \), you can solve for \( \lambda \) to find the expected rate of failures. This process is essential because determining \( \lambda \) allows you to calculate other statistics, such as the proportion of events (copier failures) expected within any period.
Finding the median is particularly helpful when planning preventative measures or understanding the scheduling of maintenance, as it indicates when a significant proportion of systems or devices need attention.
Cumulative Distribution Function
The cumulative distribution function (CDF) of an exponential distribution describes the probability that the time until an event occurs is less than a specified value. For an exponential distribution, the CDF is expressed as \( F(t) = 1 - e^{-\lambda t} \). Here, \( t \) is the time variable, and \( \lambda \) is the rate parameter.
The CDF gives you a cumulative probability value rather than just a snapshot like the PDF. When dealing with the maintenance of copiers, the CDF can be used to determine the likelihood that a machine will require servicing within a given time period, such as the first year.
In the exercise, \( \lambda \) has already been determined to be \( \ln(2)/2 \), so the probability of a copier failing within the first year is \( P(T < 1) = 1 - e^{-\ln(2)/2} \). Simplifying this gives approximately 0.2929, indicating a 29.29% chance that a copier will fail within one year.
Understanding the CDF helps in making informed decisions about maintenance schedules and resource allocations based on the expected timing of failures.

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