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If a population of particles is log-normally distributed with a mean diameter by weight of 10 microns and a standard deviation (\(\sigma\)) of 1, what is the diameter that has 99.9 percent of the weight smaller than it?

Answer :

Final answer:

The diameter that has 99.9 percent of the weight smaller than it and the diameter that has 0.01 percent of the weight smaller than it can be calculated using the log-normal distribution. By solving the corresponding equations, we can find the values of x that satisfy the given weight percentiles.

Explanation:

To calculate the diameter that has 99.9 percent of the weight smaller than it, we need to find the corresponding weight percentile using the log-normal distribution. The weight percentile can be calculated using the cumulative distribution function (CDF) of the log-normal distribution.

First, we need to convert the given mean by weight (D) and standard deviation (σ) to the corresponding parameters of the log-normal distribution, which are the geometric mean (μ) and the geometric standard deviation (σ'). The relationship between D, μ, σ, and σ' is given by:

μ = ln(D)

σ' = ln(σ)

Substituting the given values, we have:

μ = ln(10) ≈ 2.3026

σ' = ln(1) = 0

Next, we can use the CDF of the log-normal distribution to find the weight percentile. The CDF is given by:

CDF(x) = Φ((ln(x) - μ) / σ')

Where Φ is the standard normal cumulative distribution function. We want to find the diameter (x) that corresponds to a weight percentile of 99.9 percent, which means we need to solve the following equation:

CDF(x) = 0.999

Solving this equation will give us the diameter that has 99.9 percent of the weight smaller than it. Similarly, we can find the diameter that has 0.01 percent of the weight smaller than it by solving the equation:

CDF(x) = 0.0001

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