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The thickness of a conductive coating in micrometers has a density function of \(600 x^{-2}\) for \(100 \mu \mathrm{m}

Short Answer

Expert verified
Mean is 109.54 μm, variance is 12.96 μm²; average cost is $54.77 per part.

Step by step solution

01

Understand the Problem

We are given a probability density function (pdf) for the thickness of a conductive coating as \( f(x) = 600x^{-2} \) in the interval \( 100 \ \mu m < x < 120 \ \mu m \). We need to find the mean and variance of this distribution and compute the average cost of the coating per part.
02

Verify Validity of the Density Function

The density function needs to integrate to 1 over its interval. Check if \( \int_{100}^{120} 600x^{-2} \, dx = 1 \).
03

Compute the Integral of the Density Function

Calculate \( \int_{100}^{120} 600x^{-2} \, dx \). This integral evaluates to \( 600 \left[-\frac{1}{x}\right]_{100}^{120} = 600 \left(-\frac{1}{120} + \frac{1}{100}\right) = 1 \). Thus, the function is a valid probability density function.
04

Calculate the Mean of the Distribution

The mean \( \mu \) of a continuous random variable is given by \( \mu = \int_{100}^{120} x f(x) \, dx = \int_{100}^{120} 600x^{-1} \, dx \). Evaluating this gives \( 600 [\ln x]_{100}^{120} = 600 (\ln 120 - \ln 100) \approx 109.54 \).
05

Calculate the Variance of the Distribution

The variance \( \sigma^2 \) is calculated as \( \sigma^2 = \int_{100}^{120} x^2 f(x) \, dx - \mu^2 \). Compute \( \int_{100}^{120} 600x^0 \, dx \), which gives \( 600[x]_{100}^{120} = 600(20) \). Variance \( \approx 600(20) - (109.54)^2 \approx 12.96 \).
06

Determine the Average Cost of the Coating

The cost per part is \( \\(0.50 \) per micrometer. Multiply the mean thickness by the cost factor: \( 109.54 \times 0.50 = \\)54.77 \).

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Key Concepts

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

Mean of a distribution
The mean of a distribution, often represented by the symbol \( \mu \), is the expected value or average of a random variable. In simple terms, it indicates where the center of the data lies.
The concept of mean is crucial as it provides a single summary number for a large set of data points. When dealing with a continuous distribution, the mean is calculated using the probability density function (pdf).
For a continuous random variable \( X \) with a pdf \( f(x) \), the mean is determined by the integral
\[ \mu = \int_{a}^{b} x f(x) \, dx \]
where \( a \) and \( b \) define the range of the random variable. Here, this calculation reveals the true average thickness of the coating over its range.
  • This involves using calculus to integrate the product of \( x \) and its density \( f(x) \) over the specified interval.
  • The resulting mean offers an estimate of the coating thickness, simplifying further analysis such as cost calculations.
Variance of a distribution
Variance is a measure of how much the values in a distribution differ from the mean. It is expressed as \( \sigma^2 \) and essentially captures the spread of the distribution. A high variance indicates a wide spread of data, while a low variance suggests the data is closely clustered around the mean.
To compute the variance of a continuous distribution, one must first find the expected value of the square of the variable and then subtract the square of the mean:
\[ \sigma^2 = \int_{a}^{b} x^2 f(x) \, dx - \mu^2 \]
This formula tells us how much the coating thickness deviates from the mean thickness on average.
  • The first integral \( \int_{a}^{b} x^2 f(x) \, dx \) calculates the expected value of \( x^2 \).
  • By subtracting \( \mu^2 \), we adjust for the central tendency, isolating the true variability of the distribution.

Understanding variance is critical in statistics and cost calculations, as it helps predict how consistent the coating thickness will be across different parts.
Cost calculation in statistics
In statistical applications, cost calculations often involve estimating the average cost associated with a particular attribute, such as the thickness of a coating. This exercise highlights the importance of mean calculations in determining financial impacts.
Given that the cost is \( \\(0.50 \) per micrometer of thickness, the average cost can be calculated by multiplying the mean thickness by the unit cost:
\[ \text{Average Cost} = \mu \times \text{Unit Cost} \]
This straightforward multiplication gives the expected cost per part, making it easier to predict expenses associated with the coating process.
  • Central to this calculation is the use of the mean value, which offers a reliable estimate of quantity (i.e., thickness) for cost estimation.
  • The average cost, calculated as \( \\)54.77 \), represents the cost of applying the coating to a single part based on the average thickness.
  • Such direct cost analyses are vital for budgeting and financial planning in manufacturing and engineering fields, where resources need precise allocation.
Knowing the average cost per part allows companies to manage budgets effectively and plan for current and future production.

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