Math Problem Statement

graph P(X = 0) = e^{-1.4} \approx 0.2466

Solution

To graph the probability P(X=0)=e1.40.2466P(X = 0) = e^{-1.4} \approx 0.2466, we need to consider it in the context of a probability distribution. The given expression suggests we're dealing with a Poisson distribution.

Steps to Graph P(X=0)P(X = 0):

  1. Poisson Distribution Formula: The probability mass function for a Poisson distribution with parameter λ\lambda is: P(X=k)=λkeλk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} Where λ\lambda is the mean rate, and kk is the number of events (in this case, k=0k = 0).

  2. Given Information:

    • P(X=0)=e1.4P(X = 0) = e^{-1.4}, which suggests λ=1.4\lambda = 1.4 for the Poisson distribution.
    • The probability for X=0X = 0 is: P(X=0)=e1.40.2466P(X = 0) = e^{-1.4} \approx 0.2466
  3. Graphing: The Poisson distribution typically shows the probabilities for different values of XX. For λ=1.4\lambda = 1.4, the graph will show the probability mass function for integer values of XX.

    • P(X=0)=0.2466P(X = 0) = 0.2466 is one point on the graph.
    • We would also calculate and plot P(X=1),P(X=2),P(X = 1), P(X = 2), \dots to get the full distribution.

Would you like me to plot the entire Poisson distribution for λ=1.4\lambda = 1.4, or would you prefer to focus on specific parts of it?

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Poisson Distribution

Formulas

Poisson distribution probability mass function: P(X = k) = (λ^k e^(-λ)) / k!

Theorems

Poisson Distribution

Suitable Grade Level

Grades 10-12