91Ó°ÊÓ

Asbestos fibers in a dust sample are identified by an electron microscope after sample preparation. Suppose that the number of fibers is a Poisson random variable and the mean number of fibers per square centimeter of surface dust is 100\. A sample of 800 square centimeters of dust is analyzed. Assume a particular grid cell under the microscope represents \(1 / 160,000\) of the sample. (a) What is the probability that at least one fiber is visible in the grid cell? (b) What is the mean of the number of grid cells that need to be viewed to observe 10 that contain fibers? (c) What is the standard deviation of the number of grid cells that need to be viewed to observe 10 that contain fibers?

Step by step solution

01

Determine the Mean Number of Fibers in the Sample

The problem states the mean number of fibers per square centimeter is 100. For a total area of 800 square centimeters, the mean number of fibers is simply calculated by multiplying the two given values:\[\lambda = 100 \times 800 = 80,000.\]

02

Compute Mean Number of Fibers in a Grid Cell

A particular grid cell under the microscope represents \(\frac{1}{160,000}\) of the sample. Therefore, the mean number of fibers per grid cell is:\[\lambda_{\text{cell}} = \frac{80,000}{160,000} = 0.5.\]

03

Calculate Probability of Zero Fibers in a Grid Cell

The probability of observing zero fibers in a Poisson distribution is given by\[P(X=0) = e^{-\lambda_{\text{cell}}} \cdot \frac{(\lambda_{\text{cell}})^0}{0!}\]Substitute \(\lambda_{\text{cell}} = 0.5\):\[P(X=0) = e^{-0.5} \approx 0.6065.\]

04

Determine Probability of At Least One Fiber in Grid Cell

The probability of at least one fiber, \(P(X \geq 1)\), is the complement of zero fibers, calculated as\[P(X \geq 1) = 1 - P(X=0) = 1 - 0.6065 = 0.3935.\]

05

Find Mean Number of Grid Cells for Observing 10 With Fibers

The number of grid cells that show at least one fiber can be modeled as a geometric distribution with probability of success \(p = 0.3935\). The mean, or expected value, of a geometric distribution to achieve 10 successes is\[E(n) = \frac{10}{p} = \frac{10}{0.3935} \approx 25.42.\]

06

Calculate Standard Deviation for Observing 10 Cells With Fibers

The standard deviation for a geometric distribution to achieve 10 successes when not considering order is given by\[\sigma(n) = \sqrt{\frac{10(1-p)}{p^2}} = \sqrt{\frac{10(1-0.3935)}{0.3935^2}} \approx 15.85.\]

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Key Concepts

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

Geometric Distribution

The geometric distribution is a probability distribution that models the number of trials needed to achieve the first success in a series of Bernoulli trials, where each trial is independent and has the same probability of success. In the context of our exercise:

• We are interested in finding out how many grid cells need to be viewed before we observe a cell with fibers. This is considered a 'success'.
• The probability of success for observing at least one fiber in each grid cell was calculated to be approximately 0.3935.

The geometric distribution provides a simple yet powerful tool to model scenarios where we're counting the trials, like checking each grid cell until finding one with fibers.

Expected Value

The expected value is the average number of trials required to achieve the first success in a geometric distribution. It gives us an idea of how many trials we might expect to perform before success.

• In our exercise, we calculated that approximately 25.42 grid cells need to be viewed to observe 10 grid cells that contain fibers.
• This value is determined by dividing the desired number of successes (10 cells with fibers) by the probability of success in a single trial (0.3935).

Understanding the expected value helps us prepare for experimental or observational tasks, like determining how many grid cells will typically contain fibers if observing under a microscope.

Standard Deviation

Standard deviation measures the amount of variation or dispersion from the expected value. It provides insight into how spread out the number of trials might be from the mean.

• In our scenario, the standard deviation was calculated to be approximately 15.85 for the number of grid cells needed to observe 10 that contain fibers.
• This indicates a significant range or variability around the mean of 25.42 grid cells.

A higher standard deviation indicates greater variability, meaning that the actual number of trials required could vary significantly from the average number predicted by the expected value.

Probability Calculation

Probability calculation involves determining the likelihood of various outcomes in experiments or observations. In this exercise:

• We calculated the probability of observing zero fibers in a grid cell using the Poisson distribution formula, resulting in approximately 0.6065.
• The probability of observing at least one fiber was derived as the complement of observing zero fibers: approximately 0.3935.

Probability calculations are foundational in statistics, giving us the ability to predict outcomes over repeated experiments or observations.