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Chapter 4: Problem 13

What is the expected value and variance for a Poisson distribution with parameter \(\mu=4.0\) ?

Short Answer

Expert verified
Expected value is 4.0 and variance is 4.0.

Step by step solution

01

Understand the Poisson Distribution

The Poisson distribution is a probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space. It is characterized by the parameter \(\mu\), which represents the average number of events.
02

Recall the Expected Value of a Poisson Distribution

For a Poisson distribution with parameter \(\mu\), the expected value (mean) is equal to \(\mu\).
03

Compute the Expected Value

Given \(\mu = 4.0\), the expected value is \(E[X] = \mu = 4.0\).
04

Recall the Variance of a Poisson Distribution

For a Poisson distribution, the variance \(Var(X)\) is also equal to the parameter \(\mu\).
05

Compute the Variance

With \(\mu = 4.0\), the variance is \(Var(X) = 4.0\).

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Key Concepts

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

Expected Value in Poisson Distribution
The expected value, also known as the mean, is a fundamental concept in probability and statistics. In the context of the Poisson distribution, which deals with the probability of a certain number of events occurring in a fixed period, the expected value is simply the parameter \( \mu \). This parameter represents the average number of occurrences in the given time or space interval.
  • If you imagine an event happening repeatedly, the expected value tells you the average number of times this event will occur.
  • For our example, when \( \mu = 4.0 \), it means on average, 4 events are expected to occur per interval.
Variance in Poisson Distribution
Variance is a measure of how spread out the numbers in your data set are around the mean. In simpler terms, it provides a sense of the data's consistency. For the Poisson distribution, an interesting fact is that the variance is equal to the expected value. This means the variance is also \( \mu \).
  • This characteristic highlights that in a Poisson distribution, the dispersion from the average number of events is as frequent as the average itself.
  • With \( \mu = 4.0 \), variance becomes \( Var(X) = 4.0 \), indicating some variability around the mean of 4 events, yet not wildly scattered.
Understanding Probability Distribution
A probability distribution defines how the values of a random variable (like the number of events) are distributed. The Poisson distribution specifically details how likely each number of events is, given \( \mu \).
  • It helps us to model and understand random events in real life. For instance, the number of customers arriving at a store within an hour could follow a Poisson distribution.
  • The distribution is often represented graphically, showing the probability of various outcomes. Peaks in the graph show the most likely outcomes.
Poisson distributions skew towards randomness, but the expected value and variance simplify predictions. By knowing both, we gain insight into what to expect and how much variability might occur.

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Most popular questions from this chapter

A study [12] of incidence rates of blindness among insulindependent diabetics reported that the annual incidence rate of blindness per year was \(0.67 \%\) among 30 - to 39 -year-old male insulin-dependent diabetics (IDDM) and 0.74\% among 30- to 39-year-old female insulin-dependent diabetics. What does cumulative incidence mean, in words, in the context of this problem?

(TABLE CAN NOT COPY) Suppose an "otitis-prone family" is defined as one in which at least three siblings of five develop otitis media in the first 6 months of life. What proportion of five-sibling families is otitis prone if we assume the disease occurs independently for different siblings in a family?

The presence of bacteria in a urine sample (bacteriuria) is sometimes associated with symptoms of kidney disease in women. Suppose a determination of bacteriuria has been made over a large population of women at one point in time and \(5 \%\) of those sampled are positive for bacteriuria. Suppose 100 women from this population are sampled. What is the probability that 3 or more of them are positive for bacteriuria?

A study was conducted among 234 people who had expressed a desire to stop smoking but who had not yet stopped. On the day they quit smoking, their carbonmonoxide level (CO) was measured and the time was noted from the time they smoked their last cigarette to the time of the CO measurement. The CO level provides an "objective" indicator of the number of cigarettes smoked per day during the time immediately before the quit attempt. However, it is known to also be influenced by the time since the last cigarette was smoked. Thus, this time is provided as well as a "corrected CO level," which is adjusted for the time since the last cigarette was smoked. Information is also provided on the age and sex of the participants as well as each participant's self-report of the number of cigarettes smoked per day. The participants were followed for 1 year for the purpose of determining the number of days they remained abstinent. Number of days abstinent ranged from 0 for those who quit for less than 1 day to 365 for those who were abstinent for the full year. Assume all people were followed for the entire year. (TABLE CAN NOT COPY) Develop a life table similar to Table 4.16, giving the number of people who remained abstinent at \(1,2, \ldots\) 12 months of life (assume for simplicity that there are 30 days in each of the first 11 months after quitting and 35 days in the 12 th month). Plot these data on the computer using either Excel or \(R\) or some other statistical package. Compute the probability that a person will remain abstinent at \(1,3,6,\) and 12 months after quitting.

A study was performed concerning medical emergencies on commercial airline flights (Peterson, et al., [16] ). A database was constructed based on calls to a medical communications center from 5 domestic and international airlines representing approximately \(10 \%\) of global passenger flight volume from January 1,2008 to October 31,2010 . There were 11,920 in-flight medical emergencies (IFM) among 7, 198,118 flights during the study period. Assume for this entire problem that there is at most 1 IFM per flight. Suppose a flight attendant works on 2 flights per day for each of 300 days per year. What is the probability that the flight attendant will encounter at least one IFM over a 1-year period?
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