Question

In a Poisson distribution, if $\frac{P(X=5)}{P(X=2)} = \frac{1}{7500}$ and $\frac{P(X=5)}{P(X=3)} = \frac{1}{500}$, then the mean of the distribution is

• A. 15
• B. 5
• C. 25
• D. 3

Hint:

N/A

Answer 1

The Correct Option is B

The problem involves finding the mean of a Poisson distribution using given probability ratios. In a Poisson distribution, the probability of a given number of events is expressed as:

\(P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}\)

Where \( \lambda \) is the mean of the distribution, and \( x \) is the number of occurrences.

Given:

• \(\frac{P(X=5)}{P(X=2)} = \frac{1}{7500}\)
• \(\frac{P(X=5)}{P(X=3)} = \frac{1}{500}\)

Step-by-step Solution:

1. Using the formula for Poisson distribution, the ratio of probabilities is: \(\frac{P(X=5)}{P(X=2)} = \frac{e^{-\lambda} \lambda^5/5!}{e^{-\lambda} \lambda^2/2!} = \frac{\lambda^3}{60}\).
2. Equating it to the given value: \(\frac{\lambda^3}{60} = \frac{1}{7500}\)
3. Simplify to find: \(\lambda^3 = \frac{60}{7500} = \frac{1}{125}\)
4. Thus, \(\lambda = \sqrt[3]{\frac{1}{125}} = \frac{1}{5}\)
5. Use the second given ratio for verification: \(\frac{P(X=5)}{P(X=3)} = \frac{e^{-\lambda} \lambda^5/5!}{e^{-\lambda} \lambda^3/3!} = \frac{\lambda^2}{20} = \frac{1}{500}\)
6. Expanding, we find: \(\lambda^2 = \frac{20}{500} = \frac{1}{25}\)
7. Computing gives: \(\lambda = \sqrt{\frac{1}{25}} = \frac{1}{5}\)

Thus, the calculated mean from both equations matches, confirming our solution:

The mean of the distribution is 5.