Finding the Standard Deviation of a Poisson Random Variable

The question asks us to find the standard deviation of a Poisson random variable X, given the condition \(2P(X = 0) = P(X = 2)\). Understanding the properties of a Poisson distribution is key to solving this problem.

A Poisson random variable X represents the number of events occurring in a fixed interval of time or space, if these events occur with a known constant mean rate and independently of the time since the last event. The probability mass function (PMF) of a Poisson distribution with mean \(\lambda\) is given by:

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

where:

• \(P(X=k)\) is the probability of observing exactly k events.
• \(\lambda\) is the average number of events in the interval (the mean).
• \(e\) is the base of the natural logarithm (approximately 2.71828).
• \(k!\) is the factorial of k.

For a Poisson distribution, the mean and variance are equal to \(\lambda\). The standard deviation is the square root of the variance, which is \(\sqrt{\lambda}\).

Using the Given Condition to Find the Mean (\(\lambda\))

We are given the condition \(2P(X = 0) = P(X = 2)\). Let's calculate the probabilities \(P(X = 0)\) and \(P(X = 2)\) using the PMF formula:

For \(k = 0\):

\(P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!}\)

Since \(\lambda^0 = 1\) and \(0! = 1\), this simplifies to:

\(P(X = 0) = \frac{e^{-\lambda} \cdot 1}{1} = e^{-\lambda}\)

For \(k = 2\):

\(P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!}\)

Since \(2! = 2 \times 1 = 2\), this becomes:

\(P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2}\)

Now, substitute these expressions back into the given condition \(2P(X = 0) = P(X = 2)\):

\(2(e^{-\lambda}) = \frac{e^{-\lambda} \lambda^2}{2}\)

We need to solve this equation for \(\lambda\). Since \(e^{-\lambda}\) is always positive (\(e^x > 0\) for any real x), we can divide both sides of the equation by \(e^{-\lambda}\):

\(2 = \frac{\lambda^2}{2}\)

Now, multiply both sides by 2:

\(2 \times 2 = \lambda^2\)

\(4 = \lambda^2\)

Taking the square root of both sides gives us \(\lambda = \pm\sqrt{4} = \pm 2\). However, the mean (\(\lambda\)) of a Poisson distribution must be a non-negative value. Therefore, we take the positive root:

\(\lambda = 2\)

So, the mean of the Poisson distribution is 2.

Calculating the Standard Deviation

The standard deviation of a Poisson random variable is \(\sqrt{\lambda}\). We found that \(\lambda = 2\).

Standard Deviation = \(\sqrt{2}\)

Let's summarize the steps:

1. Identify the distribution as Poisson and recall its PMF, mean, variance, and standard deviation properties.
2. Use the PMF to write expressions for \(P(X=0)\) and \(P(X=2)\) in terms of \(\lambda\).
3. Substitute these expressions into the given equation \(2P(X = 0) = P(X = 2)\).
4. Solve the resulting equation for \(\lambda\).
5. Calculate the standard deviation using the value of \(\lambda\).

The standard deviation of the Poisson random variable X is \(\sqrt{2}\).

Revision Table: Poisson Distribution Properties

This table summarizes the key properties of a Poisson distribution with mean \(\lambda\).

Additional Information: Understanding Poisson Distribution

The Poisson distribution is a discrete probability distribution that models the probability of a given number of events occurring in a fixed interval. It is often used to model rare events. Examples include:

• The number of phone calls received by a call center per hour.
• The number of defects in a manufactured product per unit area.
• The number of car accidents at a specific intersection per week.
• The number of customers arriving at a store per minute.

A key characteristic of the Poisson distribution is that its mean and variance are equal. This property was crucial in our calculation, as finding the mean (\(\lambda\)) directly gave us the variance, from which the standard deviation is easily derived.

The condition \(2P(X = 0) = P(X = 2)\) provided specific information about the probabilities of observing 0 and 2 events, allowing us to determine the unique parameter \(\lambda\) for this particular Poisson distribution.