Assume that X is a Poisson random variable with = 1.5. Calculate P (X1) Hint THE POISSON DISTRIBUTION For a Poisson random variable X, the probability of x successes over a given interval of time or space is O 0.3347 O 0.2510 P(X=X)= for x = 0, 1, 2, ..., where is the mean number of successes and e≈ 2.718 is the base of the natural logarithm. O 0.4422 e
Assume that X is a Poisson random variable with = 1.5. Calculate P (X1) Hint THE POISSON DISTRIBUTION For a Poisson random variable X, the probability of x successes over a given interval of time or space is O 0.3347 O 0.2510 P(X=X)= for x = 0, 1, 2, ..., where is the mean number of successes and e≈ 2.718 is the base of the natural logarithm. O 0.4422 e
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Poisson Distribution Problem
**Problem Statement:**
Assume that \( X \) is a Poisson random variable with \( \mu = 1.5 \).
Calculate \( P(X - 1) \).
**Hint:**
#### The Poisson Distribution
For a Poisson random variable \( X \), the probability of \( x \) successes over a given interval of time or space is:
\[
P(X = x) = \frac{e^{-\mu} \mu^x}{x!}
\]
for \( x = 0, 1, 2, \ldots \), where \( \mu \) is the mean number of successes and \( e \approx 2.718 \) is the base of the natural logarithm.
**Answer Options:**
- \( 0.3347 \)
- \( 0.2510 \)
- \( 0.4422 \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b5c41b5-9371-4990-b74c-6e4519fbbd36%2F164e3473-88b7-43ec-b5d0-2e4477dafdf9%2Fy00dlyl_processed.png&w=3840&q=75)
Transcribed Image Text:### Poisson Distribution Problem
**Problem Statement:**
Assume that \( X \) is a Poisson random variable with \( \mu = 1.5 \).
Calculate \( P(X - 1) \).
**Hint:**
#### The Poisson Distribution
For a Poisson random variable \( X \), the probability of \( x \) successes over a given interval of time or space is:
\[
P(X = x) = \frac{e^{-\mu} \mu^x}{x!}
\]
for \( x = 0, 1, 2, \ldots \), where \( \mu \) is the mean number of successes and \( e \approx 2.718 \) is the base of the natural logarithm.
**Answer Options:**
- \( 0.3347 \)
- \( 0.2510 \)
- \( 0.4422 \)
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