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The mean life of a component under a Weibull distribution is an important measure, as it gives an average lifespan that we might expect for that component. In a Weibull distribution, mean life can be calculated even with varying rates of failure over time.
We use this formula to find the mean life: \[ \mu = \delta \cdot \Gamma\left(1 + \frac{1}{\beta}\right) \]This formula is essential and incorporates two parameters:

• Shape parameter \( \beta \): It describes how the failure rate changes over time.
• Scale parameter \( \delta \): It stretches or compresses the scale of time.

For our pump, where \( \beta = 2 \) and \( \delta = 700 \), we plug these into the formula along with \( \Gamma(1.5) = 0.88623 \). The resulting mean life \( \mu \) is approximately 620.36 hours. This calculation shows how long on average a pump lasts, assuming the characteristics modeled by this distribution.

Variance, in the context of the Weibull distribution, describes the spread or variability of the life of components from the mean, offering insight into the consistency of the lifespan.
To find the variance \( \sigma^2 \), use this formula:\[\sigma^2 = \delta^2 \left(\Gamma\left(1 + \frac{2}{\beta}\right) - \left(\Gamma\left(1 + \frac{1}{\beta}\right)\right)^2\right) \]We deploy our Weibull shape \( \beta = 2 \) and scale \( \delta = 700 \), using \( \Gamma(2) = 1 \) for calculations. The formula evaluates as:

• Calculate \( \Gamma\left(1 + \frac{1}{\beta}\right) \) as before: \( 0.88623 \).
• Compute \( 490000 \times(1 - 0.7854) \).

This results in a variance of about 105154 hours², suggesting a moderate spread of life expectancies around the mean of 620.36 hours.

Understanding the probability that a component lasts beyond its mean life involves exploring the cumulative distribution function (CDF). This function, specifically in the Weibull context, can determine how likely a pump will outperform its average lifespan.
For probability, the focus shifts to: \[ P(X > \mu) = 1 - F(\mu) \]Where \( F(\mu) \) is the CDF at the mean, calculated as:\[ F(x) = 1 - \exp\left(-\left(\frac{x}{\delta}\right)^\beta\right) \]For \( \mu \approx 620.36 \):

• Compute the ratio \( \left(\frac{620.36}{700}\right)^2 \) and use it within the exponential function.
• Find \( 1 - \exp(-0.7854) \), yielding roughly 0.5444.

Thus, the probability \( P(X > \mu) \approx 0.4556 \). This yields an insight that there's nearly a 45.56% chance a pump could last beyond its mean lifespan, providing a significant perspective on reliability beyond average performance.