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Chapter 17: Problem 12

The random variable \(X\) is Poisson distributed with mean \(\mu\) and satisfies \(P(x=3)=P(x=0)+P(x=1).\) a) Find the value of \(\mu,\) correct to four decimal places. b) For this value of \(\mu\) evaluate \(P(2 \leqslant x \leqslant 4)\)

Short Answer

Expert verified
a) \(\mu \approx 1.8892\); b) \(P(2 \leq x \leq 4) \approx 0.5138\).

Step by step solution

01

Understanding the Poisson Probability Mass Function

The probability mass function for a Poisson distributed random variable with mean \( \mu \) is given by:\[P(X = k) = \frac{\mu^k \cdot e^{-\mu}}{k!}\]where \( k \) is the number of occurrences (which can be 0, 1, 2, ...).
02

Set up the equation from the given condition

Given condition is \( P(X = 3) = P(X = 0) + P(X = 1) \). Write down the expressions for each using the Poisson PMF:\[P(X = 3) = \frac{\mu^3 \cdot e^{-\mu}}{3!}\]\[P(X = 0) = \frac{\mu^0 \cdot e^{-\mu}}{0!} = e^{-\mu}\]\[P(X = 1) = \frac{\mu^1 \cdot e^{-\mu}}{1!} = \mu e^{-\mu}\]So, substitute these into the condition:\[\frac{\mu^3 \cdot e^{-\mu}}{6} = e^{-\mu} + \mu e^{-\mu}\]
03

Simplify the equation

Factor out \( e^{-\mu} \) from both sides:\[\frac{\mu^3}{6} = 1 + \mu\]
04

Solve the equation for \( \mu \)

Solve \( \mu^3 = 6 + 6\mu \) using numerical methods or by trial and error for \( \mu \). The solution gives \( \mu \approx 1.8892 \).
05

Calculate \( P(2 \leq X \leq 4) \)

Now calculate the probabilities for \( X = 2, 3, 4 \) using \( \mu = 1.8892 \).\[P(X = 2) = \frac{(1.8892)^2 \cdot e^{-1.8892}}{2!}\]\[P(X = 3) = \frac{(1.8892)^3 \cdot e^{-1.8892}}{3!}\]\[P(X = 4) = \frac{(1.8892)^4 \cdot e^{-1.8892}}{4!}\]Calculate each probability and sum them:\[P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)\]
06

Final calculations and answer

Calculate based on the previous formulas:\[P(X=2) \approx 0.2662, \quad P(X=3) \approx 0.1683, \quad P(X=4) \approx 0.0793\]Then,\[P(2 \leq X \leq 4) \approx 0.2662 + 0.1683 + 0.0793 = 0.5138\]

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Key Concepts

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

Probability Mass Function
Understanding the probability mass function (PMF) is essential when dealing with Poisson distributions. The PMF gives us the probability of a random variable taking a specific value. For a Poisson-distributed random variable, the PMF is represented by the formula:\[ P(X = k) = \frac{\mu^k \cdot e^{-\mu}}{k!} \]Here,
  • \(X\) is a random variable indicating the number of occurrences.
  • \(\mu\) is the mean value, representing the average number of occurrences.
  • \(k\) is a non-negative integer indicating a specific number of occurrences.
  • The term \(e^{-\mu}\) allows the probability to decrease as \(\mu\) increases.
  • The factorial \(k!\) aids in adjusting how probability diminishes as the count grows.
This formula helps us understand how the likelihood of different outcomes, such as 0, 1, or 3 occurrences, can be computed, given a certain mean.
Random Variable
In probability and statistics, a random variable is a variable that takes on different values based on the outcome of a random phenomenon. In a Poisson distribution, the random variable \(X\) represents the number of events occurring within a fixed interval of time or space. The variable can take on values such as 0, 1, 2, 3, and so on.Random variables can be classified into discrete and continuous, but in the Poisson distribution, we focus on discrete random variables. This is because:
  • They represent countable events, like the number of emails received in an hour.
  • The events are independent; occurrence of one does not affect another.
The Poisson distribution specifically models events that occur at a constant mean rate. Our problem utilizes a random variable \(X\) to find out the probability relationships based on a set mean.
Mean Value
In the context of Poisson distribution, the mean value \(\mu\) is more than just an average; it signifies the expected number of occurrences over a specific interval. Typically, for Poisson distributions:
  • The mean value indicates the intensity or rate of event occurrences.
  • It is equal to the variance of the distribution, highlighting its impact on spread as well.
When solving problems with Poisson distributions, finding the appropriate mean value \(\mu\) is crucial in calculating probabilities. For instance, with our problem's setup, the relationship \(P(X=3) = P(X=0) + P(X=1)\) provided us a path to determine \(\mu\). We used trial and error or computational methods to find \(\mu \approx 1.8892\), which was pivotal in solving the exercise and evaluating the probabilities for different ranges of \(X\).

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Most popular questions from this chapter

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The discrete random variable \(X\) has probability function given by $$P(x)=\left\\{\begin{array}{cc} \left(\frac{1}{4}\right)^{x-1} & x=2,3,4,5,6 \\ k & x=7 \\ 0 & \text { otherwise } \end{array}\right.$$ where \(k\) is a constant. Determine the value of \(k\) and the expected value of \(X .\)
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