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A random variable is a fundamental concept in probability and statistics. It is essentially a variable that can take on different values, each with a certain probability. In the context of the given exercise, the time to first failure of an engine valve is considered a random variable, denoted by \( Y \). This particular variable can take any value that represents thousands of hours until the valve fails.
The behavior of the random variable \( Y \) is modeled using a probability density function (pdf), which defines how probabilities are distributed over the values of the random variable. The provided pdf, \( f(y) = 2y e^{-y^2} \) for \( y \geq 0 \), illustrates the probability that the valve will fail after a certain amount of time. Understanding how this function behaves across its domain is crucial to analyzing the likelihood of different outcomes.
Critical properties of a pdf include:

• Non-negativity: The function must be non-negative for all values in its domain, which means that probabilities cannot be negative.
• Total probability is 1: The area under the pdf across its domain must equal 1, ensuring that one of the possible outcomes happens with certainty.

The mean of a random variable, also known as the expected value, provides an average measure of where the outcomes of the random variable are centered. In simpler terms, it's what you might "expect" on average if you could observe the outcomes many times. The mean is calculated using an integral involving the random variable's pdf.
In the problem setup, the mean of the valve failure time \( Y \) is determined using the integral \( E(Y) = \int_{0}^{\infty} y \, 2y e^{-y^2} \, dy \). By changing variables and integrating, it was shown that the mean is \( \frac{\sqrt{\pi}}{2} \), providing insight into the average timeframe in thousands of hours that one can expect the valve to last.

• The mean is particularly useful for comparison across different probability distributions.
• It helps in planning and preparation, such as maintenance schedules that rely on the typical behavior of the system.

Probability calculations are essential when analyzing how likely certain outcomes are, given a probability density function. This might include finding out how probable it is for an event to occur within a certain range. In the exercise, several probability calculations were performed.
For instance, to assess the likelihood that a valve lasts at least a certain time before failing, one can integrate the pdf from that time to infinity. This calculation uses the expression \( P(Y \geq 2) = \int_{2}^{\infty} 2y e^{-y^2} \, dy = e^{-4} \), determining the probability that a valve lasts at least 2000 hours.
Other probability calculations might involve finding probabilities that at least one out of several valves fails using principles like independence and complements.
Understanding probability calculations enables:

• Making informed decisions and risk assessments based on likelihoods.
• Performing predictive analysis for real-world planning applications.

The interquartile range (IQR) is a measure of statistical dispersion, representing the range within which the central 50% of data points lie. It is the difference between the 75th percentile (third quartile) and the 25th percentile (first quartile) of a distribution.
In the context of the given problem, determining the IQR involves calculating the values of \( Y \) that correspond to the 25% and 75% cumulative probabilities. Specifically, the problem solved for these percentiles using the cumulative distribution function derived from the pdf:

• 25th percentile (\( y_{0.25} \)): \( y_{0.25} = \sqrt{-\ln(0.75)} \)
• 75th percentile (\( y_{0.75} \)): \( y_{0.75} = \sqrt{-\ln(0.25)} \)

The IQR is then the difference \( y_{0.75} - y_{0.25} \).
Using the IQR is beneficial because:

• It provides a robust measure of variability that isn't sensitive to extreme values (outliers).
• It's useful for identifying the spread of the data around the median.
• Can be visualized effectively using box plots in data analysis.