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Let \(x\) be a Poisson random variable with mean \(\mu=2.5 .\) Use Table 2 in Appendix I to calculate these probabilities: a. \(P(x \geq 5)\) b. \(P(x<6)\) c. \(P(x=2)\) d. \(P(1 \leq x \leq 4)\)

Short Answer

Expert verified
Based on the given Poisson random variable with a mean of 2.5, the following probabilities were calculated: a. The probability of x being greater than or equal to 5, P(x ≥ 5), is 0.1089. b. The probability of x being less than 6, P(x < 6), is 0.9579. c. The probability of x being equal to 2, P(x = 2), is 0.2565. d. The probability of x being between 1 and 4, inclusive, P(1 ≤ x ≤ 4), is 0.8090.

Step by step solution

01

Understanding the Poisson Probability Distribution

The Poisson distribution is defined by the probability mass function: \(P(x=k) = \frac{e^{-\mu}\mu^k}{k!}\). Here, we have \(\mu = 2.5\), and we want to find various probabilities using the table provided.
02

Finding P(x >= 5)

To find the probability \(P(x \geq 5)\), we need to find the sum of the probabilities of x = 5, 6, 7, ..., which can be written as \(P(x\geq5) = 1 - P(x<5)\). Now, we sum the probabilities for x = 0, 1, 2, 3, and 4. From the table, we find \(P(x=0) = 0.0821\), \(P(x=1) = 0.2052\), \(P(x=2) = 0.2565\), \(P(x=3) = 0.2137\), \(P(x=4) = 0.1336\) \(P(x \geq 5) = 1 - (0.0821 + 0.2052 + 0.2565 + 0.2137 + 0.1336)\) \(P(x \geq 5) = 1 - 0.8911\) \(P(x \geq 5) = 0.1089\)
03

Finding P(x < 6)

For this probability, we sum the probabilities for x = 0, 1, 2, 3, 4, and 5: \(P(x < 6) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5)\) \(P(x < 6) = 0.0821 + 0.2052 + 0.2565 + 0.2137 + 0.1336 + 0.0668\) (found from the table) \(P(x < 6) = 0.9579\)
04

Finding P(x = 2)

This probability has already been found in previous steps as \(P(x=2) = 0.2565\).
05

Finding P(1

To find this probability, we sum the probabilities for x = 1, 2, 3, and 4: \(P(1 \leq x \leq 4) = P(x=1) + P(x=2) + P(x=3) + P(x=4)\) \(P(1 \leq x \leq 4) = 0.2052 + 0.2565 + 0.2137 + 0.1336\) \(P(1 \leq x \leq 4) = 0.8090\) In conclusion, our probabilities are: a. \(P(x \geq 5) = 0.1089\) b. \(P(x < 6) = 0.9579\) c. \(P(x = 2) = 0.2565\) d. \(P(1 \leq x \leq 4) = 0.8090\)

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is a key component of understanding probability distributions, especially discrete ones like the Poisson distribution. In the case of a Poisson distribution, the PMF helps us find the probability that a random variable takes a specific number of events happening in a fixed interval of time or space. The formula for the PMF in a Poisson distribution is given by:\[P(x = k) = \frac{e^{-\mu}\mu^k}{k!}\]Where:
  • \(k\) represents the number of occurrences we want to find the probability for.
  • \(\mu\) is the average number of occurrences (the mean of the distribution).
  • \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
  • \(k!\) denotes the factorial of \(k\), which is the product of all positive integers up to \(k\).
To utilize the PMF, we replace \(k\) with our desired value in the formula. This allows us to calculate the probability of that specific number of occurrences. It's important to remember that the sum of all probabilities in a distribution will equal 1.
Expected Value
The concept of the Expected Value (or mean) in a Poisson distribution is crucial for determining the central tendency of the data. The Expected Value tells us the average number of occurrences in a given dataset over a fixed interval.In the context of the Poisson distribution, the Expected Value is very straightforward; it is simply the parameter \(\mu\) itself. So, if you have \(\mu = 2.5\), then the average or expected number of times our event occurs in an interval is 2.5.The Expected Value is a measure of the 'center' of the distribution, giving us an idea of what to anticipate in terms of event frequency. It's what you would expect to happen if an experiment could be repeated an infinite number of times under the same conditions.
Probability Calculation
Calculating probabilities using a Poisson distribution involves using the formula from the Probability Mass Function. It can sometimes also require expressing cumulative probabilities or combining individual probabilities for ranges of outcomes.For instance:- For \(P(x \geq 5)\), we use the complementary probability: \(1 - P(x < 5)\). This means first calculating the sum of probabilities from 0 to 4, then subtracting from 1.- For calculating \(P(x < 6)\), we sum the probabilities for each event from 0 up through 5.These calculations may reference a table of pre-calculated Poisson probabilities like the one mentioned in the exercise. Just remember to substitute your values into the PMF formula and assess what's required, whether an exact probability or a range of outcomes.Understanding such calculations is vital because it allows you to predict event frequencies either in times of need or for planning in scenarios with stochastic processes.

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Most popular questions from this chapter

A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let \(x\) equal the number of women chosen to fill the two positions. a. Write the formula for \(p(x)\), the probability distribution of \(x\). b. What are the mean and variance of this distribution? c. Construct a probability histogram for \(x\).

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