Question

If a drop of water is examined under a microscope, the number \(x\) of a specific type of bacteria present has been found to have a Poisson probability distribution. Suppose the maximum permissible count per water specimen for this type of bacteria is five. If the mean count for your water supply is two and you test a single specimen, is it likely that the count will exceed the maximum permissible count? Explain.

Short answer

The probability of a single specimen having a count that exceeds the maximum permissible count (5) is approximately 1.66%. Since this probability is quite low, it is not considered likely that a single specimen will have a count exceeding the maximum permissible count.

Step-by-step solution

Define Poisson probability distribution

The count of bacteria follows a Poisson probability distribution. The probability mass function (PMF) for a Poisson distribution can be defined as: \(P(k; \lambda) = \frac{\lambda^{k}e^{-\lambda}}{k!}\) where \(k\) is the number of events (bacteria in this case), \(\lambda\) is the mean count and \(e\) is the Euler's number (approximately 2.718). Here, we have \(\lambda\) = 2 (mean count for the water supply).

Calculate the probability of the number of bacteria exceeding the maximum permissible count

We need to find the probability of the count exceeding 5. In other words, we need to calculate the probability of the count being 6 or more: \(P(X > 5) = 1 - P(X \leq 5)\) To find \(P(X \leq 5)\), we will sum the probabilities for each \(k\) count from 0 to 5: \(P(X \leq 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)\) Now we will calculate each of these probabilities using the Poisson probability mass function: \(P(0) = \frac{2^{0}e^{-2}}{0!} = e^{-2}\) \(P(1) = \frac{2^{1}e^{-2}}{1!} = 2e^{-2}\) \(P(2) = \frac{2^{2}e^{-2}}{2!} = 2e^{-2}\) \(P(3) = \frac{2^{3}e^{-2}}{3!} = \frac{4}{3}e^{-2}\) \(P(4) = \frac{2^{4}e^{-2}}{4!} = \frac{1}{3}e^{-2}\) \(P(5) = \frac{2^{5}e^{-2}}{5!} = \frac{2}{15}e^{-2}\) Now add these probabilities: \(P(X \leq 5) = e^{-2} + 2e^{-2} + 2e^{-2} + \frac{4}{3}e^{-2} + \frac{1}{3}e^{-2} + \frac{2}{15}e^{-2}\) \(P(X \leq 5) \approx 0.983436391\) Then we can find the probability of the count exceeding 5: \(P(X > 5) = 1 - P(X \leq 5) \approx 1 - 0.983436391 \approx 0.016563609\)

Determine whether the count exceeding the maximum is considered likely

Now that we have calculated the probability of the specimen count exceeding the maximum permissible count (i.e., \(P(X > 5) \approx 0.0166\)), we can determine if it's considered likely. In general, an event is considered likely if its probability is significantly greater than 0. In this case, the probability of \(P(X > 5)\) is quite low, approximately 1.66%. Therefore, it is not likely that a single specimen will have a count that exceeds the maximum permissible count (5).

Key concepts

Probability Mass Function

The probability mass function (PMF) is a fundamental concept in statistics, used for describing the probability of each possible outcome in a discrete random variable scenario. It quantifies the likelihood that a discrete random variable is exactly equal to some value. In the context of our exercise, the PMF is the key tool that helps determine the likelihood of observing a certain number of bacteria in a water specimen.

For a Poisson distribution, the PMF is particularly useful when dealing with events that occur independently and with a known average rate, symbolized by \( \lambda \). The Poisson PMF is given by the formula:
\[ P(k; \lambda) = \frac{\lambda^{k}e^{-\lambda}}{k!} \]
where \( k \) represents the number of events (or bacteria count), \( \lambda \) is the average number of events (the mean bacteria count), and \( e \) is Euler's number. To calculate the likelihood of various bacteria counts, you would plug different values of \( k \) into this formula. By understanding and applying the PMF, students can grasp how to assess the probabilities of different outcomes for a given mean rate of occurrence.

Euler's Number

Euler's number, denoted as \( e \), is an irrational and transcendental number that is approximately equal to 2.71828. It is a cornerstone of mathematics, particularly in calculus and complex analysis, but it also appears in the realm of probability and statistics - notably in our exercise through the Poisson distribution.

Euler's number arises naturally when dealing with continuous growth or decay processes, like interest calculations in finance or radioactive decay in physics. In the Poisson probability mass function, \( e \) is inversely proportional to the likelihood of observing more events. This is because as \( k \) increases, \( e^{-\lambda} \) shrinks, reflecting the diminished likelihood of higher event counts. When explaining Euler's number to students, emphasize its role as the base of the natural logarithm, making it integral to exponential growth models and the Poisson distribution formula used in our exercise.

Statistical Significance

Statistical significance is a measure of whether the results of an experiment or study are likely to be true and not occurred by chance. It indicates the reliability of an effect or difference found in a dataset; in simple terms, it's used to determine if the observed data provides enough evidence to assert a conclusion about a population parameter.

For the problem given, statistical significance would help us decide if the probability of a water specimen having a bacteria count that exceeds five is just by random chance or indicative of a real issue with the water supply. A statistically significant outcome is typically one in which the probability of the event occurring by chance is low, often below a threshold value such as 5% (the common \( p < 0.05 \) criterion). In the example, a specimen count exceeding five happens with a probability of approximately 1.66%, which is below the 5% threshold. Thus, it would be considered statistically significant, indicating it is unlikely that such a high bacteria count is due to chance alone.