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Chapter 6: Problem 9

The random variable \(X\) follows a Poisson process with the given value of \(\lambda\) and \(t\). Assuming \(\lambda=0.07\) and \(t=10\), compute (a) \(P(4)\) (b) \(P(X<4)\) (c) \(P(X \geq 4)\) (d) \(P(4 \leq X \leq 6)\) (e) \(\mu_{X}\) and \(\sigma_{X}\)

Short Answer

Expert verified
(a) 0.00496 (b) 0.752 (c) 0.248 (d) 0.005608 (e) \mu_X = 0.7, \sigma_X = 0.83666

Step by step solution

01

Define the Poisson Distribution

The Poisson distribution is defined as \[ P(X = k) = \frac{{e^{-\text{rate}} \text{rate}^k}}{{k!}} \] where the rate or mean \[ \text{rate} = \text{lambda} \times t \].Given \[ \text{lambda} = 0.07 \] and \[ t = 10 \], we have \[ \text{rate} = \text{lambda} \times t = 0.07 \times 10 = 0.7 \].
02

Compute P(4)

To find \[ P(X = 4) \], use the Poisson formula: \[ P(X = 4) = \frac{{e^{-0.7} \times 0.7^4}}{{4!}} \].Calculating, \[ P(X = 4) = \frac{{e^{-0.7} \times 0.2401}}{{24}} \ = \frac{{0.496585 \times 0.2401}}{{24}} \ = 0.00496 \].
03

Compute P(X

To find \[ P(X < 4) \], we sum the probabilities from \[ P(X = 0) \] to \[ P(X = 3) \]. \[ P(X < 4) = \text{Sum} \bigg(\frac{{e^{-0.7} \times 0.7^0}}{{0!}}, \frac{{e^{-0.7} \times 0.7^1}}{{1!}}, \frac{{e^{-0.7} \times 0.7^2}}{{2!}}, \frac{{e^{-0.7} \times 0.7^3}}{{3!}}\bigg) = 0.496585 \big(1 + 0.7 + 0.245 + 0.0571667\big)\ = 0.752 \].
04

Compute P(X 鈮� 4)

To find \[ P(X \geq 4) \], we use \[ P(X \geq 4) = 1 - P(X < 4) \]. Using our previous result for \[ P(X < 4) \], we get \[ P(X \geq 4) = 1 - 0.752 = 0.248 \].
05

Compute P(4 \leq X \leq 6)

Between 4 and 6 implies \[ P(4) + P(5) + P(6) \].We already have \[ P(X = 4) \]. Now calculate \[ P(X = 5) = \frac{{e^{-0.7} \times 0.7^5}}{{5!}} \] and \[ P(X = 6) = \frac{{e^{-0.7} \times 0.7^6}}{{6!}} \ = 0.00058 \ and \0.000068 \]. Therefore, \[ P(4 \leq X \leq 6) = 0.00496 + 0.00058 + 0.000068 = 0.005608 \].
06

Compute the Mean and Standard Deviation

The mean (\[ \mu_X \]) of a Poisson distribution is rate. The standard deviation (\[ \sigma_X \]) is the square root of rate. Thus,\[ \mu_X = 0.7 \ and \ \sigma_X = \sqrt{0.7} = 0.83666 \].

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Key Concepts

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

Poisson Distribution
The Poisson distribution is a probability distribution that describes events occurring independently within a fixed interval of time or space. It is particularly useful for modeling rare events. The key parameter here is the rate (\text{rate}), which is the product of \text{lambda} (average rate of occurrence) and t (time period).
  • Formula: \( P(X = k) = \frac{{e^{-\text{rate}} \text{rate}^k}}{{k!}} \)
  • The mean \text{rate} for our problem is calculated as 0.07 (lambda) \( \times \) 10 (t), giving us 0.7.
Knowing the rate enables us to compute individual probabilities for any given number of events within the interval.
Probability Computation
To solve any Poisson problem, compute probabilities for specific occurrences using the Poisson formula. Below, we break down the calculations for given examples.
  • Calculate \( P(X = 4) \):
    \( P(X = 4) = \frac{{e^{-0.7} \times 0.7^4}}{{4!}} \).
    By simplification, \( P(X = 4) = 0.00496 \).

  • Calculate \( P(X < 4) \):
    Sum probabilities of \( P(X = 0), P(X = 1), P(X = 2), \) and \( P(X = 3) \).
    \( P(X < 4) = 0.752 \).

  • Calculate \( P(X \geq 4) \):
    \( P(X \geq 4) = 1 - P(X < 4) \).
    \( P(X \geq 4) = 0.248 \).

  • Calculate \( P(4 \leq X \leq 6) \):
    Sum \( P(X = 4), P(X = 5), \) and \( P(X = 6) \).
    \( P(4 \leq X \leq 6) = 0.005608 \).
Understanding these computations helps you handle different types of questions involving events under a Poisson distribution.
Mean and Standard Deviation
For a Poisson distributed random variable, the mean (\( \mu_X \)) and the standard deviation (\( \sigma_X \)) offer key insights.
  • The mean \( \mu_X \) is simply the rate of occurrence: \( \lambda \times t \).
  • For our data, \( \mu_X = 0.7 \).

  • The standard deviation \( \sigma_X \) is the square root of the rate.
  • For our data, \( \sigma_X = \sqrt{0.7} \approx 0.83666 \).
These values characterize the central tendency and variability in the number of occurrences within the given interval.

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