0.2021 Question 14 (30 points) ✓ Saved The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters $\alpha=2$, $\beta=3$. What is the variance of X? O 1.926 O 0.766 O 2.662 O 0.632
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The probability density function (PDF) of a Weibull distribution is given by: $f(x; \alpha, \beta) = \frac{\beta}{\alpha} \left(\frac{x}{\alpha}\right)^{\beta-1} e^{-(x/\alpha)^\beta}$ for $x \ge 0$ The mean of a Weibull distribution is given by: $E[X] = \alpha Show more…
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