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 every 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 \(r\) (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 fewer than four times during the 30 -minute observation period.

Step by step solution

01

Determine the Frequency of Grooming

Given that the frequency of initiating grooming is 1.7 for every 10 minutes, in a 30-minute observation period, the expected number of grooming initiations would be calculated by multiplying the rate by 3. Thus, \[\lambda = 1.7 \times 3 = 5.1\] where \(\lambda\) is the average rate of occurrence over the interval.

02

Identify Poisson Distribution Fit

The Poisson distribution is an appropriate model when events occur independently over a fixed period of time and the average rate (\(\lambda\)) is known. Here, grooming initiations satisfy these criteria with a fixed period (30 minutes) and known rate (\(\lambda=5.1\)), so the Poisson distribution is a good fit.

03

Write the Poisson Probability Formula

The probability mass function for a Poisson distribution is\[P(r) = \frac{\lambda^r e^{-\lambda}}{r!}\]where \(r\) is the number of occurrences, \(\lambda\) is the average rate, and \(e\) is the base of the natural logarithm, approximately equal to 2.718.

04

Calculate Probabilities for Four, Five, and Six Groomings

Using \(\lambda = 5.1\), compute the probabilities:- For \(r = 4\):\[P(4) = \frac{5.1^4 e^{-5.1}}{4!} \approx 0.172\]- For \(r = 5\):\[P(5) = \frac{5.1^5 e^{-5.1}}{5!} \approx 0.174\]- For \(r = 6\):\[P(6) = \frac{5.1^6 e^{-5.1}}{6!} \approx 0.148\]

05

Probability of Grooming Four or More Times

Calculate the cumulative probability for \(r \geq 4\) by subtracting the cumulative probability for \(r < 4\) from 1:\[P(r \geq 4) = 1 - (P(0) + P(1) + P(2) + P(3))\]Calculate each:- \(P(0) \approx 0.006\)- \(P(1) \approx 0.031\)- \(P(2) \approx 0.078\)- \(P(3) \approx 0.132\)Then the cumulative probability:\[P(r \geq 4) \approx 1 - (0.006 + 0.031 + 0.078 + 0.132) = 1 - 0.247 \approx 0.753\]

06

Probability of Fewer than Four Groomings

To find \(P(r < 4)\), sum the probabilities for \(r = 0, 1, 2, 3\):\[P(r < 4) = P(0) + P(1) + P(2) + P(3) \approx 0.006 + 0.031 + 0.078 + 0.132 = 0.247\]

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Key Concepts

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

Probability Mass Function

One of the key concepts of the Poisson distribution is the Probability Mass Function (PMF). This function helps us understand the probability of a certain number of events occurring within a fixed interval. In this exercise, the event is the initiation of social grooming by a river otter over 30 minutes. The Poisson PMF is given by:\[P(r) = \frac{\lambda^r e^{-\lambda}}{r!}\]Here,

• \(r\) is the number of times the event occurs,
• \(\lambda\) is the average number of event occurrences in the given time period, and
• \(e\) is the base of the natural logarithm, approximately 2.718.

The PMF gives probabilities for specific values of \(r\), like 4, 5, or 6 times. This is crucial for tasks like calculating whether an otter grooms a certain number of times.

Expected Value

The Expected Value in a Poisson distribution is synonymous with \(\lambda\), representing the average rate of occurrence of the event over the observed time period. It is calculated as the average number of times an event is likely to occur. In this case, grooming initiations are spread over 30 minutes, starting from a given rate of 1.7 every 10 minutes.For a 30-minute interval, we calculate:\[\lambda = 1.7 \times 3 = 5.1\]This \(\lambda\) value tells us that, on average, there are 5.1 grooming initiations in 30 minutes. The concept of expected value is foundational as it helps to summarize the distribution with a single number characterized by a simple average.

Cumulative Probability

Cumulative Probability refers to the probability that the random variable is less than or equal to a particular value (for example, \(r\leq 3\)) or more than that (like \(r \geq 4\)). This is particularly useful for understanding the probability of a range of outcomes rather than just one outcome.To find cumulative probability, sometimes called the "tail probability," for \(r \geq 4\), you use:\[P(r \geq 4) = 1 - (P(0) + P(1) + P(2) + P(3))\]This applies the chance of the complementary event \(r < 4\), calculated as the sum of probabilities for \(r = 0, 1, 2, 3\), subtracted from 1. This approach provides a comprehensive view, letting us explore the likelihood of any set of occurrences as well as specific intervals.

Random Variable

In statistics, a Random Variable is a variable whose values depend on the outcomes of a random phenomenon. In the context of this exercise, the random variable \(r\) represents the number of times an otter initiates social grooming in 30 minutes.Random variables can be discrete or continuous. When dealing with the Poisson distribution, the random variable is discrete, taking on integer values such as 0, 1, 2, etc.The Poisson distribution models the random variable \(r\) as it helps to describe situations where events occur independently, and the average event rate (\(\lambda\)) is constant over time. Understanding the nature and role of the random variable is pivotal in applying the Poisson distribution to real-world problems, allowing us to make informed predictions based on the probabilities calculated.