📘 Poisson Distribution Numerical Examples
Author: Bindeshwar Singh Kushwaha
Institution: PostNetwork Academy
Example 1: Truck Arrivals
The number of heavy trucks arriving at a railway station follows a Poisson distribution with an average of 2 arrivals per hour.
Find:
- (a) Probability that no truck arrives
- (b) Probability that at least two trucks arrive
Let \( \lambda = 2 \). Poisson PMF:
\[
P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \dots
\]
(a) \( P(X = 0) = e^{-2} \approx 0.1353 \)
(b)
\[
P(X \geq 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – (0.1353 + 0.2706) = 0.5941
\]
Example 2: Serum Reaction
Probability of bad reaction = 0.001, total individuals = 1500.
\( \lambda = np = 1.5 \)
- (i) Exactly 3 suffer:
\[
P(X = 3) = \frac{e^{-1.5} \cdot 1.5^3}{3!} = \frac{e^{-1.5} \cdot 3.375}{6} \approx 0.1255
\]
- (ii) More than 2 suffer:
\[
P(X > 2) = 1 – P(X \leq 2)
\]
\[
P(X \leq 2) = e^{-1.5}(1 + 1.5 + 1.125) = e^{-1.5}(3.625) \approx 0.8097
\]
\[
P(X > 2) \approx 1 – 0.8097 = 0.1903
\]
Example 3: Moments of Poisson Distribution
If mean \( \lambda = 1.44 \):
- Variance: \( \lambda = 1.44 \)
- Third central moment: \( \mu_3 = 1.44 \)
- Fourth central moment:
\[
\mu_4 = 3\lambda^2 + \lambda = 3(1.44)^2 + 1.44 = 7.66
\]
Example 4: Solve for \( \lambda \) from a Condition
Given \( P(X = 1) = 2P(X = 2) \)
\[
\frac{e^{-\lambda} \lambda}{1!} = 2 \cdot \frac{e^{-\lambda} \lambda^2}{2!} \Rightarrow \lambda = \lambda^2 \Rightarrow \lambda(\lambda – 1) = 0
\]
Valid solution: \( \lambda = 1 \)
Mean = Variance = 1
Example 5: Sum of Poisson Variables
If \( X \sim \text{Poisson}(1) \), \( Y \sim \text{Poisson}(2) \), and they are independent, then:
\( X + Y \sim \text{Poisson}(3) \)
Find \( P(X + Y < 2) \):
\[
P(W < 2) = P(W = 0) + P(W = 1) = e^{-3} + 3e^{-3} = 4e^{-3} \approx 0.1992
\]
Example 6: Mine Accident Probability
Given \( p = \frac{1}{1400} \), \( n = 350 \Rightarrow \lambda = 0.25 \)
\[
P(X \geq 1) = 1 – P(X = 0) = 1 – e^{-0.25} \approx 0.2212
\]
There is a 22.12% chance of at least one fatal accident.
Example 7: Validity Check
Mean = 6 → \( \lambda = 6 \)
Standard deviation = 2 → Variance = 4 → \( \lambda = 4 \)
Contradiction → Invalid Poisson assumption.
Example 8: Equating Probabilities
Given \( P(X = 1) = P(X = 2) \), solve:
\[
\frac{e^{-\lambda} \lambda}{1!} = \frac{e^{-\lambda} \lambda^2}{2!} \Rightarrow \lambda = \frac{\lambda^2}{2} \Rightarrow \lambda(\lambda – 2) = 0
\Rightarrow \lambda = 2
\]
Then:
\[
P(X = 0) = e^{-2} \approx 0.1353,\quad
P(X = 4) = \frac{e^{-2} \cdot 2^4}{4!} = \frac{16}{24}e^{-2} \approx 0.0902
\]
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