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Chapter 5: Problem 47

Bacteria in Water Samples If a drop of water is placed on a slide and examined under a microscope, the number \(x\) of a particular 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

Expert verified
Answer: The probability of the bacteria count exceeding the maximum permissible count of 5 is approximately 55.78%.

Step by step solution

01

Understand the Poisson Distribution Formula

In a Poisson distribution, the probability mass function is given by the formula: \(P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}\) where \(X\) is the number of events, \(k\) is the number of successes we are testing for, \(\lambda\) is the mean number of events, and \(e \approx 2.718\) is the base of the natural logarithm. Here, \(X\) is the bacteria count and \(\lambda\) is the mean count, which is given as \(2\). We want to find the probability that the count will exceed the maximum permissible count of \(5\).
02

Calculate the Probability of Bacteria Count Less Than or Equal to 5

We will first find the probability that the bacteria count is less than or equal to \(5\). This can be calculated using the Poisson distribution formula for each count from \(0\) to \(5\), and summing up those probabilities: \(P(X \leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)\) Now, we will plug in the values of \(k\) and \(\lambda\) to calculate each probability and sum them up: \(P(X \leq 5) = \frac{e^{-2} \cdot 2^0}{0!} + \frac{e^{-2} \cdot 2^1}{1!} + \frac{e^{-2} \cdot 2^2}{2!} + \frac{e^{-2} \cdot 2^3}{3!} + \frac{e^{-2} \cdot 2^4}{4!} + \frac{e^{-2} \cdot 2^5}{5!}\) \(P(X \leq 5) \approx 0.4422\)
03

Find the Probability of Bacteria Count Exceeding 5

Now that we have found the probability of the bacteria count being less than or equal to \(5\), we can find the probability of it exceeding \(5\) by subtracting the calculated probability from \(1\): \(P(X > 5) = 1 - P(X \leq 5)\) \(P(X > 5) = 1 - 0.4422\) \(P(X > 5) \approx 0.5578\) The probability of the bacteria count exceeding the maximum permissible count of \(5\) is approximately \(55.78\%\). This suggests that it is likely that a single specimen from the water supply will have a bacteria count that exceeds the maximum permissible limit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Bacteria Count with Poisson Distribution
When you examine a water sample under a microscope, the number of a particular type of bacteria present is referred to as the bacteria count. In many cases, such as this, the bacteria count follows a Poisson distribution.

The Poisson distribution is used to represent the number of events occurring within a fixed interval of time or space. It is particularly useful for modeling rare events, such as finding bacteria in a particular drop of water.
  • The parameter \( \lambda \) (lambda) signifies the mean number of times an event occurs in a given interval鈥攈ere, it is the average count of bacteria鈥攁nd is given as \( 2 \) in this context.
  • In a Poisson distribution, the events are independent, and the average rate (mean) of occurrence is constant.
Understanding how the bacteria count follows a Poisson distribution is crucial for predicting the likelihood of observing a particular count like 5 bacteria in a sample.
Probability Calculation for Bacteria Count
To calculate the likelihood of a certain bacteria count, we use the Poisson probability mass function. This function helps determine the probability of observing an exact number of events, \( k \), given the average rate of occurrence, \( \lambda \), and is expressed as:
\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]For our exercise, we need to calculate the probability of a bacteria count being \(5\) or fewer.
  • First, calculate probabilities for each count from \(0\) to \(5\) using the equation.
  • These individual probabilities are then summed up, providing the cumulative probability \( P(X \leq 5) \).
The calculated value, approximately \(0.4422\), tells us the probability of having five or fewer bacteria in the sample. An essential step is comprehending that the sum of probabilities provides insight into the expected range of observations.
Understanding Maximum Permissible Count
The concept of a maximum permissible count refers to the threshold number that is considered acceptable or safe by some criteria. In this context, the "maximum permissible count" is \(5\) bacteria per water specimen.

We want to understand if the actual count in a sample is notably higher than this permissible count.
After calculating, we find that the probability of the count exceeding \(5\) bacteria is \(0.5578\). This calculation is straightforward: we subtract the probability of \( X \leq 5 \) from \(1\).
  • This result (\(55.78\%\)) indicates a more than 50% probability that a water specimen will exceed the permissible bacteria count.
  • A high percentage warns that the water supply may frequently surpass the safety standard, signaling a potential health risk.
By understanding these calculations, you can assess whether actions should be taken to address the bacteria levels in the water supply.

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Most popular questions from this chapter

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