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Chapter 9: Problem 31

Waiting Time The waiting time \(t\) for service in a store is exponentially distributed with a mean of 5 minutes. (a) Find the probability density function of the random variable \(t\) (b) Find the probability that \(t\) is within one standard deviation of the mean.

Short Answer

Expert verified
The probability density function is \(f(t) = \frac{1}{5} e^{-\frac{1}{5}t}\) for \(t \geq 0\), and \(0\) otherwise. The probability that \(t\) lies within one standard deviation of the mean is \(1 - e^{-2}\).

Step by step solution

01

Find the probability density function

A random variable, \(t\), is said to follow an exponential distribution with parameter \(\lambda = \frac{1}{\mu}\), where \(\mu\) is the mean of the distribution. Therefore, the probability density function (PDF) of an exponential distribution is given by: \(f(t) = \lambda e^{-\lambda t}\) for \(t \geq 0\), and \(0\) otherwise. In this case, the mean (\(\mu\)) is 5 minutes, thus \(\lambda = \frac{1}{5}\). So, the PDF of \(t\), denoted by \(f(t)\), is given by: \(f(t) = \frac{1}{5} e^{-\frac{1}{5}t}\) for \(t \geq 0\), otherwise \(0\).
02

Find the standard deviation

The standard deviation of an exponential distribution is the same as the mean. So, in this case, the standard deviation is also 5 minutes.
03

Find the probability that \(t\) is within one standard deviation of the mean

We need to find the probability that \(t\) falls between 0 and 10 (since 5-0 and 5+5 gives our range). So we need to integrate the PDF from 0 to 10. The result is given by: \(\int_{0}^{10} \frac{1}{5} e^{-\frac{1}{5}t} dt =1 - e^{-2}\).

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Key Concepts

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

Probability Density Function
In the context of an exponential distribution, the probability density function (PDF) is a crucial concept.
It helps us understand how probabilities are distributed over time.
For a random variable that is exponentially distributed, the PDF helps us determine the likelihood of certain time intervals occurring in a process like waiting time.

The exponential distribution is characterized by the parameter \(\lambda\), which is defined as the reciprocal of the mean \(\mu\) of the distribution.
That means \(\lambda = \frac{1}{\mu}\).
The formula for the probability density function of an exponential distribution is:
\[f(t) = \lambda e^{-\lambda t}\] for \(t \geq 0\).
This equation indicates that the probability of the event quickly decreases as time increases.

In our example, the mean \(\mu\) is 5 minutes, which implies \(\lambda = \frac{1}{5}\).
Therefore, the PDF becomes \(f(t) = \frac{1}{5} e^{-\frac{1}{5}t}\).
This function helps us find the probabilities associated with different waiting times.
Standard Deviation
The standard deviation of a probability distribution tells us about the amount of variability or spread in the data.
For an exponential distribution, there's a unique relationship between the mean and standard deviation.

Particularly, the standard deviation \(\sigma\) of an exponential distribution is exactly equal to its mean \(\mu\).
This is a rare feature among probability distributions.
Therefore, if the mean is 5 minutes, the standard deviation is also 5 minutes.
This means that the time variation in waiting times is spread evenly around the mean value, making it easier to anticipate future occurrences.

This information is extremely useful when estimating probabilities for time intervals, including events happening within a specific duration from the mean.
Mean of Distribution
The mean of a distribution represents the average or expected value.It's a central concept in probability and statistics, providing a summary point for the entire distribution.
In the case of exponential distribution, it offers insights into the average waiting time or the average time until an event occurs.

For an exponential distribution, the mean \(\mu\) not only reflects the typical value but also plays a direct role in determining the rate parameter \(\lambda\).
Given by \(\lambda = \frac{1}{\mu}\), this relationship shows that a larger mean results in a smaller rate parameter, implying slower incidents or events.

In our exercise example, which deals with waiting time, the mean is 5 minutes.
This value signifies that, on average, a customer waits approximately 5 minutes for service.
All other calculations, like determining the PDF or making probability estimates, rely on this mean value.
Hence, it acts as the backbone for effectively interpreting and managing the exponential distribution.

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Most popular questions from this chapter

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