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The exponential distribution is a continuous probability distribution used to model the time between independently occurring events. In this context, it describes the diameter of drilled holes that are affected by factors like machinery vibrations and tool wear. The probability density function (PDF) for an exponential distribution is expressed as \( f(x) = eta e^{-eta x} \), where \( eta \) is the rate parameter. This parameter helps us understand the frequency of events (like faulty drill outcomes) and is key in calculating both the mean and variance.

In the given problem, the exponential distribution's PDF is defined as \( f(x) = 10e^{-10(x-5)} \). This equation indicates that our rate parameter \( \lambda \) is 10. Understanding the PDF is crucial because it forms the foundation for determining probabilities and further statistical measures such as the mean and variance.

Calculating the mean and variance of an exponential distribution allows us to summarize the data with measures of central tendency and spread. For an exponential distribution, the mean \( \mu \) is calculated using the formula \( \mu = \frac{1}{\lambda} \). In our case, given \( \lambda = 10 \), the mean is \( \mu = \frac{1}{10} = 0.1 \).

However, because the random variable \( x \) is affected by an additional factor of 5 (in the PDF \( f(x) = 10e^{-10(x-5)} \)), we add 5 to account for the shift, resulting in a mean of \( 5.1 \) millimeters.

• Exponential distributions have an inherent variance, which describes data variability. The variance \( \sigma^2 \) is given by \( \sigma^2 = \frac{1}{\lambda^2} \).
• Given \( \lambda = 10 \), the variance is \( \sigma^2 = \frac{1}{10^2} = 0.01 \).

This variance implies the spread and deviations around the mean hole diameter.

The cumulative distribution function (CDF) provides a measure of the probability that a random variable like the hole diameter takes a value less than or equal to a specific value. It reflects the area under the PDF curve up to that value.

For exponential distributions, the CDF is calculated as \( F(x) = 1 - e^{-\lambda(x-5)} \). To find the CDF for a hole diameter greater than 5, we evaluate \( F(5.1) \). Here, it is calculated as:

• \( F(5.1) = 1 - e^{-1} \)
• \( e^{-1} \approx 0.3679 \)

The CDF helps assert that the probability of seeing a value up to \( 5.1 \) millimeters is \( 1 - e^{-1} \). The result indicates the probability of acceptable diameter outcomes.

Integration by parts is a calculus technique that helps in integrating products of functions. It is a valuable tool when deriving functions such as the CDF from the exponential PDF.

The basic formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). It is instrumental in breaking down complex exponentials into integrable components.

• Given our PDF, \( f(x) = 10e^{-10(x-5)} \), integration by parts allows us to convert this into an easier form to handle expected value and further extends to find cumulative probabilities.
• Using it efficiently requires picking \( u \) and \( dv \) such that the process simplifies integration, leading us to the CDF calculation.

This technique ultimately simplifies problems requiring integration, especially in probabilistic contexts where exponentials are involved.