Problem 20 - An Introduction to Mathematical Statistics and Its Applications

Question

Suppose a radioactive source is metered for two hours, during which time the total number of alpha particles counted is four hundred eighty-two. What is the probability that exactly three particles will be counted in the next two minutes? Answer the question two ways-first, by defining \(X\) to be the number of particles counted in two minutes, and second, by defining \(X\) to be the number of particles counted in one minute.

Step-by-Step Solution

Answer

The probability that exactly three particles will be counted in the next two minutes is calculated using \( \lambda = 8.03 \), and the probability that exactly three particles will be counted in the next one minute is calculated using \( \lambda = 4.01667 \).

Step 1 - Calculate Rate for 2 minutes

Find the rate \( \lambda \) for two minutes. As given, 482 particles are counted in 120 minutes, so the average rate for two minutes is (482/120)*2 = 8.03.

Step 2 - Calculate Probability for 3 particles in 2 minutes

Use the formula for Poisson distribution to find out the probability of exactly three particles being counted in the next two minutes. The probability mass function of Poisson's distribution is given by, \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\), where \( \lambda \) is the rate per interval, k is the number of occurrences. Plugging in \( \lambda = 8.03 \) and \( k = 3 \), we obtain \( P(X=3) = \frac{e^{-8.03}(8.03)^3}{3!} \).

Step 3 -Calculate Rate for 1 minute

Next, find the rate \( \lambda \) for one minute. Given, 482 particles are counted in 120 minutes, so the average rate for one minute is 482/120 = 4.01667.

Step 4 - Calculate Probability for 3 particles in 1 minute

Finally, use the formula for Poisson distribution to find out the probability of exactly three particles being counted in the next one minute. The probability mass function of Poisson's distribution is given by, \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\), where \( \lambda \) is the rate per interval and k is the number of occurrences. Plugging in \( \lambda = 4.01667 \) and \( k = 3 \), we obtain \( P(X=3) = \frac{e^{-4.01667}(4.01667)^3}{3!} \).

Key Concepts

Poisson Distribution Formula

The Poisson distribution is a statistical formula used to model the probability of a given number of events happening within a fixed interval of time or space, provided these events occur with a known constant mean rate and independently of the time since the last event. The formula for the Poisson probability mass function is expressed as
\( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \),
where \( \lambda \) is the average rate of events per interval, \( k \) is the actual number of events that occur, \( e \) is the base of the natural logarithm (approximately equal to 2.71828), and \( k! \) represents the factorial of \( k \) which is the product of all positive integers less than or equal to \( k \).

For example, if a radioactive source emits alpha particles at an average rate of 8.03 particles every two minutes, to find the probability of exactly 3 particles being detected in the next two minutes, you would substitute these values into the Poisson formula.

Radioactive Decay

Radioactive decay is a random process by which an unstable atomic nucleus loses energy by radiation. A particular form of radioactive decay is alpha decay, where an alpha particle, which consists of two protons and two neutrons, is emitted from the nucleus. The rate at which these particles are emitted can often be well-modeled by the Poisson distribution, as the emissions are typically random and independent events occurring at a consistent rate. In our exercise, the emission of alpha particles by the radioactive source was metered over a period to determine the probability of a specified emission event using the Poisson distribution.

Probability Mass Function

In probability theory, a probability mass function (PMF) assigns probabilities to the possible values for a discrete random variable. It is a function that gives us the probability that a discrete random variable is exactly equal to some value. The sum of all probabilities for all possible values always equals 1, ensuring that the PMF represents a valid probability distribution. For the Poisson distribution, the PMF is particularly useful because it directly expresses the probability of observing exactly \( k \) events in a fixed interval, given the average event rate \( \lambda \).

Statistical Rate Calculations

Statistical rate calculations involve determining the average number of occurrences of an event within a specified interval. In the context of Poisson distribution, the rate \( \lambda \) is central since it's the average frequency at which the events occur and is used in the probability calculations. To calculate this rate, you typically divide the total number of events observed by the total length of the observation period. For instance, if a radioactive material emits 482 particles in 120 minutes, the rate is calculated as 482/120 for a one-minute interval, and this average rate is essential when working with the Poisson formula to predict the probability of events.