91Ó°ÊÓ

The expected value, often referred to as the mean, is a fundamental concept in probability that helps us understand the average or expected outcome of a random variable in the long run. For the Weibull distribution, the expected value \( E(X) \) is calculated using the formula:\[ E(X) = \alpha \Gamma\left(1 + \frac{1}{\beta}\right) \]where \( \alpha \) and \( \beta \) are the scale and shape parameters of the Weibull distribution, respectively, and \( \Gamma \) represents the gamma function. In our specific problem, where \( \alpha = 2 \) and \( \beta = 3 \), we computed\( E(X) \) by calculating \( \Gamma(1.333) \approx 0.902745 \), leading to \( E(X) \approx 1.80549 \) hundreds of hours.

• The expected value gives a central tendency or a typical value for the distribution.
• It helps us understand how long, on average, the vacuum tube is expected to last.

Variance measures the spread of a set of numbers. It tells us how much the values of a random variable differ from the expected value. For the Weibull distribution, variance \( V(X) \) is given by the formula:\[ V(X) = \alpha^2 \left( \Gamma\left(1 + \frac{2}{\beta}\right) - [\Gamma\left(1 + \frac{1}{\beta}\right)]^2 \right) \]Here, \( \alpha \) and \( \beta \) are the distribution's scale and shape parameters.In the exercise, by substituting \( \alpha = 2 \) and \( \beta = 3 \), we found the variance to be approximately 0.35098.

• Variance gives us an idea of the variability or consistency of the vacuum tube lifetime.
• Lower variance indicates that the lifetimes are closely packed around the expected value, reducing unpredictability.

The cumulative distribution function (CDF) is a key tool in probability theory that gives the probability of a random variable taking on a value less than or equal to a certain point. For the Weibull distribution, the CDF is expressed as:\[ F(x) = 1 - e^{-(x/\alpha)^\beta} \]where \( \alpha \) and \( \beta \) are the scale and shape parameters.In the problem, this formula is used to calculate the probabilities that the lifetime of the vacuum tube is less than or equal to specific values.

• To find \( P(X \leq 6) \), we calculate \( F(6) \), resulting in approximately 1 due to the very small value of \( e^{-27} \).
• For \( P(1.5 \leq X \leq 6) \), the calculation involves \( F(6) - F(1.5) \), resulting in approximately 0.655, meaning there's a 65.5% chance that the tube lasts between 1.5 and 6 hundred hours.

The gamma function is an important part of probability and statistics, especially in distributions like the Weibull distribution. It's a continuous extension of the factorial function, applicable to all complex numbers except negative integers.The gamma function, \( \Gamma(n) \), is related to factorial by the equation \( \Gamma(n) = (n-1)! \) when \( n \) is a positive integer.It is used to compute the expected value and variance in the Weibull distribution.

• For non-integer values, such as 1.333 in this problem, the Gamma function can be calculated using interpolation or specialized computational tools.
• It simplifies calculations involving complex, non-integer exponentials that arise in life data modeling like the Weibull distribution.