91Ó°ÊÓ

In his doctoral thesis, L. A. Beckel (University of Minnesota, 1982 ) studied the social behavior of river otters during the mating season. An important role in the bonding process of river otters is very short periods of social grooming. After extensive observations, Dr. Beckel found that one group of river otters under study had a frequency of initiating grooming of approximately \(1.7\) for each 10 minutes. Suppose that you are observing river otters for 30 minutes. Let \(r=0,1,2, \ldots\) be a random variable that represents the number of times (in a 30-minute interval) one otter initiates social grooming of another. (a) Explain why the Poisson distribution would be a good choice for the probability distribution of \(r\). What is \(\lambda\) ? Write out the formula for the probability distribution of the random variable \(x\) (b) Find the probabilities that in your 30 minutes of observation, one otter will initiate social grooming four times, five times, and six times. (c) Find the probability that one otter will initiate social grooming four or more times during the 30-minute observation period. (d) Find the probability that one otter will initiate social grooming less than four times during the 30-minute observation period.

Step by step solution

01

Understand the Poisson Distribution

The Poisson distribution is appropriate for counting the number of times an event happens in a fixed interval of time or space. It is characterized by the parameter \( \lambda \), which represents the average number of occurrences in that interval. Because grooming is a rare, discrete event over a short time interval, and given the frequency is uniform, the Poisson distribution fits well here.

02

Determine the Parameter \( \lambda \)

Since the grooming frequency is 1.7 per 10 minutes, for a 30-minute interval, \( \lambda = 1.7 \times 3 = 5.1 \).

03

Write the Poisson Probability Formula

The probability of observing \( r \) events in an interval is given by the Poisson probability formula: \( P(r; \lambda) = \frac{e^{-\lambda} \lambda^r}{r!} \). Here, \( \lambda = 5.1 \).

04

Calculate \( P(4) \)

Calculate the probability for \( r = 4 \) events using the formula: \( P(4; 5.1) = \frac{e^{-5.1} \times 5.1^4}{4!} \). This calculates to approximately 0.1725.

05

Calculate \( P(5) \)

Using the formula, calculate \( P(5; 5.1) = \frac{e^{-5.1} \times 5.1^5}{5!} \). This is approximately 0.1755.

06

Calculate \( P(6) \)

For \( r = 6 \), \( P(6; 5.1) = \frac{e^{-5.1} \times 5.1^6}{6!} \). This is approximately 0.1491.

07

Calculate Probability of 4 or More

The probability that \( r \geq 4 \) is 1 minus the probability that \( r < 4 \): \( P(r \geq 4) = 1 - (P(0) + P(1) + P(2) + P(3)) \). Compute each probability using the formula, then sum them to subtract from 1.

08

Calculate Probability of Less than 4

For \( r < 4 \), sum the probabilities: \( P(0), P(1), P(2), P(3) \) using the Poisson formula for each and add them together.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Random Variable

In the context of Poisson distribution, a random variable is a concept that embodies the number of times an event occurs within a specific time frame or space interval. For the river otters' grooming situation, the random variable \( r \) represents the number of social grooming initiations by an otter in a 30-minute observation period. Random variables can take on various values, in this case, \( r \) can be 0, 1, 2, and so on, depending on how many grooming sessions are observed. When we define \( r \) as a random variable, we're talking about a function that assigns a number to each outcome of a chance experiment—in this case, the number of grooming occurrences.

Probability Distribution

A probability distribution assigns a probability to each possible value of a random variable. The Poisson distribution is one type of probability distribution that is particularly useful for situations where events occur independently, at a constant average rate, within a fixed interval of time or space. For the grooming example, the Poisson probability distribution will provide us with the likelihood of observing different numbers of grooming events, such as 0, 1, 2, or more times in 30 minutes. This distribution tells us how these probabilities are distributed over possible values of our random variable \( r \). To calculate these probabilities, we use the Poisson probability formula, which we'll discuss in more detail in the probability calculation section.

Lambda Parameter

The lambda parameter, often denoted as \( \lambda \), is a crucial element of the Poisson distribution. It signifies the average number of events that occur in a given time frame or space interval. In the otters' grooming study, the frequency of grooming is 1.7 times per 10 minutes. Since we are interested in a 30-minute observation period, \( \lambda \) is calculated as \( 1.7 \times 3 = 5.1 \). This means the expected average number of times an otter initiates social grooming is 5.1 times in 30 minutes.\( \lambda \) helps define the shape of the probability distribution by determining how spread out or concentrated the occurrence of events will be over the interval.

Probability Calculation

Calculating probabilities using the Poisson distribution involves the application of its probability mass function, which is given by:\[P(r; \lambda) = \frac{e^{-\lambda} \cdot \lambda^r}{r!}\]This formula allows us to find the probability that a random variable takes on a specific value \( r \). For example, to find the probability that there are exactly 4 grooming sessions, we plug \( r = 4 \) and \( \lambda = 5.1 \) into the formula:\[P(4; 5.1) = \frac{e^{-5.1} \cdot 5.1^4}{4!} \approx 0.1725\]We apply the same process for \( r = 5 \) and \( r = 6 \). After finding these individual probabilities, we can also determine cumulative probabilities, such as the probability of observing four or more sessions by summing complementary probabilities for \( r < 4 \) and subtracting from 1.