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

To solve this problem, we need to use the properties of the Poisson distribution. In a Poisson distribution, the probability of observing \(X = k\) is given by the formula:

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

where \(\lambda\) is the mean of the distribution, \(k\) is the number of occurrences, and \(e\) is the base of the natural logarithm.

We are given two ratios:

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

Substituting the Poisson's probability formula into these ratios, we get:

1. \(\frac{\frac{\lambda^5 e^{-\lambda}}{5!}}{\frac{\lambda^2 e^{-\lambda}}{2!}} = \frac{1}{7500}\)
2. \(\frac{\frac{\lambda^5 e^{-\lambda}}{5!}}{\frac{\lambda^3 e^{-\lambda}}{3!}} = \frac{1}{500}\)

Simplifying these expressions:

1. \(\frac{\lambda^5}{\lambda^2} \cdot \frac{2!}{5!} = \frac{1}{7500}\)
2. \(\frac{\lambda^5}{\lambda^3} \cdot \frac{3!}{5!} = \frac{1}{500}\)

This reduces to:

1. \(\frac{\lambda^3}{60} = \frac{1}{7500}\) → \(\lambda^3 = \frac{60}{7500} = \frac{1}{125}\)
2. \(\frac{\lambda^2}{20} = \frac{1}{500}\) → \(\lambda^2 = \frac{20}{500} = \frac{1}{25}\)

From the above equations, we have:

1. \(\lambda^3 = \frac{1}{125}\)
2. \(\lambda^2 = \frac{1}{25}\)

Let's solve for \(\lambda\) using these two equations. It is clear that:

Taking square root from the second equation:

\(\lambda = \sqrt{\frac{1}{25}} = \frac{1}{5}\)

This indicates that \(\lambda = 5\).

Therefore, the mean of the distribution is 5.

Thus, the correct answer is 5.