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Chapter 7: Problem 7

Mr. Smith works on a temporary basis. The mean length of each job he gets is three months. If the amount of time he spends between jobs is exponentially distributed with mean 2, then at what rate does \(\mathrm{Mr}\). Smith get new jobs?

Short Answer

Expert verified
Mr. Smith gets new jobs at a rate of 6 jobs per year.

Step by step solution

01

Convert the average time between jobs to rate parameter

According to the problem, the average time Mr. Smith spends between jobs is 2 months, and this follows an exponential distribution. The exponential distribution has a mean equal to the reciprocal of its rate parameter (\(\lambda\)). Thus, we need to find the rate parameter (\(\lambda\)) by taking the reciprocal of the mean. Mean = 1 / \(\lambda\) Plugging in the given mean(2) in the formula, we'll get; 2 = 1 / \(\lambda\)
02

Solve for the rate parameter \(\lambda\)

Now we just need to solve the equation for the rate parameter, \(\lambda\): 2 = 1 / \(\lambda\) Multiplying both sides of the equation by \(\lambda\), we'll get: 2\(\lambda\) = 1 Now, we just need to divide both sides of the equation by 2 to get \(\lambda\): \(\lambda\) = 1 / 2
03

Write the rate in jobs per year

We know that \(\lambda\) represents the rate that Mr. Smith gets new jobs, and the result 1/2 means 1 new job every 2 months. But to get the rate per year, we need to multiply it by the total months in a year. Rate per year = \(\lambda \times\) (number of months in a year) There are 12 months in a year, so: Rate per year = \(\frac{1}{2} \times 12\)
04

Calculate the rate per year

Now just solve for the rate per year by multiplying the rate we found earlier by the total number of months in a year: Rate per year = \(\frac{1}{2} \times 12 = 6\) So, Mr. Smith gets new jobs at a rate of 6 jobs per year.

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Key Concepts

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

Probability Models
In the world of mathematics and statistics, a probability model is a mathematical representation of a random phenomenon. These models are constructed to analyze and predict the likelihood of various outcomes. A classic example of a probability model is the flipping of a coin, where there are two possible outcomes: heads or tails, each with a probability of 0.5, assuming the coin is fair.

Probability models can be used for more complex situations as well, such as predicting the weather, assessing risk in insurance, or even estimating the time between job assignments, as in the case study of Mr. Smith's employment. These models allow us to quantify uncertainty and make informed decisions based on the structure and parameters of the model.
Exponential Distribution
The exponential distribution is a type of probability model that is especially useful for modeling the time between events in a continuous-time Poisson process. This particular distribution is defined by a single parameter: the rate parameter \(\lambda\). It provides us with a way to model events that are independent and occur at a constant average rate across time.

In our scenario with Mr. Smith, the exponential distribution describes the time he spends between jobs. When the mean time between jobs is known (2 months), the rate parameter \(\lambda\) is simply the reciprocal of this mean (\(\lambda = \frac{1}{\text{mean}}\)). This parameter \(\lambda\) is essential because it characterizes the entire distribution, giving us the average rate at which events (in this case, job acquisitions) are occurring.
Mean Time Between Events
The mean time between events is a critical concept in various probability distributions, including the exponential distribution. It represents the average time that elapses from one event to the next. This average time can be interpreted as the reciprocal of the rate parameter in the exponential distribution (\(\text{mean} = \frac{1}{\lambda}\)).

Understanding the mean time between events is vital for planning and forecasting in fields like supply chain management, customer service, and workforce management. For Mr. Smith, knowing the mean time between jobs helps him manage his finances and expectations regarding work continuity. Moreover, converting the mean time to a rate parameter and then to an annual rate, as demonstrated in the provided exercise, serves to translate abstract concepts into practical measures applicable to real-life situations, like how frequently one can expect to be employed in a given year.

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Most popular questions from this chapter

To prove Equation ( \(7.24)\), define the following notation: \(X_{i}^{j} \equiv\) time spent in state \(i\) on the \(j\) th visit to this state; \(N_{i}(m) \equiv\) number of visits to state \(i\) in the first \(m\) transitions In terms of this notation, write expressions for (a) the amount of time during the first \(m\) transitions that the process is in state \(i ;\) (b) the proportion of time during the first \(m\) transitions that the process is in state \(i\) Argue that, with probability 1 , (c) \(\sum_{j=1}^{N_{i}(m)} \frac{X_{i}^{j}}{N_{i}(m)} \rightarrow \mu_{i}\) as \(m\) (d) \(\mathrm{N}_{i}(m) / m \rightarrow \pi_{i} \quad\) as \(m \rightarrow \infty\). (e) Combine parts (a), (b), (c), and (d) to prove Equation (7.24).

Suppose that the interarrival distribution for a renewal process is Poisson distributed with mean \(\mu .\) That is, suppose $$ P\left\\{X_{n}=k\right\\}=e^{-\mu} \frac{\mu^{k}}{k !}, \quad k=0,1, \ldots $$ (a) Find the distribution of \(S_{n}\). (b) Calculate \(P\\{N(t)=n\\}\).

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2)\) ? (b) exponentially distributed with mean \(1 ?\)

Events occur according to a Poisson process with rate \(\lambda\). Any event that occurs within a time \(d\) of the event that immediately preceded it is called a \(d\) -event. For instance, if \(d=1\) and events occur at times \(2,2.8,4,6,6.6, \ldots\), then the events at times \(2.8\) and \(6.6\) would be \(d\) -events. (a) At what rate do \(d\) -events occur? (b) What proportion of all events are \(d\) -events?

In 1984 the country of Morocco in an attempt to determine the average amount of time that tourists spend in that country on a visit tried two different sampling procedures. In one, they questioned randomly chosen tourists as they were leaving the country; in the other, they questioned randomly chosen guests at hotels. (Each tourist stayed at a hotel.) The average visiting time of the 3000 tourists chosen from hotels was \(17.8\), whereas the average visiting time of the 12,321 tourists questioned at departure was \(9.0 .\) Can you explain this discrepancy? Does it necessarily imply a mistake?
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