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The "Dependence of Vintage ASTM A7 Steel" study describes the distribution of manganese within A7 steel specimens. The nearest-neighbor distance (NND) of manganese particles along longitudinal planes in A7 steel follows a Weibull distribution with approximate parameter values α = 1.18 and β = 21.61.

(a) What is the probability of observing an NND between 20 and 40μm? Less than 20μm? More than 40μm?

(b) What are the mean and standard deviation of this distribution?

(c) What is the median of this distribution?

User Xtlc
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Final answer:

The probability, mean, standard deviation, and median of a Weibull distribution with parameters α = 1.18 and β = 21.61 require the use of a statistical software for precise calculations.

Step-by-step explanation:

The study described uses Weibull distribution to model the nearest-neighbor distance (NND) of manganese particles in ASTM A7 steel. With parameters α (shape) = 1.18 and β (scale) = 21.61, various probabilities and statistical measures can be calculated using the Weibull probability density function (pdf) and cumulative distribution function (cdf).

Probability Calculations

To find the probability of observing an NND between 20 and 40μm, we use the cdf to calculate the probability of being less than 40μm and subtract the probability of being less than 20μm. To find probabilities of being less than or greater than a specific value, we directly use the cdf or 1 - cdf respectively.

Mean and Standard Deviation

The mean of a Weibull distribution is given by β Γ(1 + 1/α), and the standard deviation is β √(Γ(1 + 2/α) - (Γ(1 + 1/α))^2), where Γ represents the gamma function.

Median of the Distribution

The median of the Weibull distribution is calculated using the formula β(ln(2))^(1/α).

Note: Calculations of probabilities, mean, standard deviation, and median would require the use of a statistical software or calculator that can handle the Weibull distribution and the gamma function (Γ).

User Mtn Pete
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