To solve this, we need to use properties of the Weibull distribution with parameters $\alpha$ (shape) and $\beta$ (scale). Given that:

• $X \sim \text{Weibull}(\alpha = 2, \beta = 3)$,
• $X$ represents the lifetime in hundreds of hours.

Part (a): $E(X)$ and $V(X)$

For a Weibull distributed random variable $X \sim \text{Weibull}(\alpha, \beta)$:

1. Expected Value $E(X)$: $E(X) = \beta \cdot \Gamma \left(1 + \frac{1}{\alpha}\right)$

2. Variance $V(X)$: $V(X) = \beta^2 \cdot \left[ \Gamma \left(1 + \frac{2}{\alpha}\right) - \left( \Gamma \left(1 + \frac{1}{\alpha}\right) \right)^2 \right]$

Where $\Gamma(\cdot)$ is the Gamma function.

Plugging in $\alpha = 2$ and $\beta = 3$:

Calculations:

1. $E(X) = 3 \cdot \Gamma \left(1 + \frac{1}{2}\right)$.

   • $\Gamma\left(1 + \frac{1}{2}\right) = \Gamma\left(\frac{3}{2}\right) \approx 0.886$.
   • So, $E(X) \approx 3 \cdot 0.886 = 2.658$.

2. $V(X) = 3^2 \cdot \left[ \Gamma\left(1 + \frac{2}{2}\right) - \left( \Gamma\left(1 + \frac{1}{2}\right) \right)^2 \right]$.

   • $\Gamma\left(1 + \frac{2}{2}\right) = \Gamma(2) = 1$.
   • Thus, $V(X) \approx 9 \cdot (1 - (0.886)^2) \approx 9 \cdot (1 - 0.785) = 9 \cdot 0.215 = 1.935$.

Answers:

• $E(X) \approx 2.658$
• $V(X) \approx 1.935$

Part (b): $P(X \leq 4)$

For a Weibull distributed random variable, the cumulative distribution function (CDF) is: $P(X \leq x) = 1 - e^{-(x / \beta)^\alpha}$

To find $P(X \leq 4)$, set $x = 4$, $\alpha = 2$, and $\beta = 3$:

$P(X \leq 4) = 1 - e^{-(4 / 3)^2}$

Calculating:

1. $(4 / 3)^2 = 1.778$.
2. $e^{-1.778} \approx 0.169$.
3. So, $P(X \leq 4) \approx 1 - 0.169 = 0.831$.

Answer:

• $P(X \leq 4) \approx 0.831$

Part (c): $P(2.5 \leq X \leq 4)$

This is calculated by finding $P(X \leq 4)$ and $P(X \leq 2.5)$, then taking the difference.

1. We already know $P(X \leq 4) \approx 0.831$.

2. For $P(X \leq 2.5)$: $P(X \leq 2.5) = 1 - e^{-(2.5 / 3)^2}$

   Calculating:

   • $(2.5 / 3)^2 = 0.694$.
   • $e^{-0.694} \approx 0.500$.
   • So, $P(X \leq 2.5) \approx 1 - 0.500 = 0.500$.

3. Now, $P(2.5 \leq X \leq 4) = P(X \leq 4) - P(X \leq 2.5) \approx 0.831 - 0.500 = 0.331$.

Answer:

• $P(2.5 \leq X \leq 4) \approx 0.331$

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Summary:

• (a) $E(X) \approx 2.658$, $V(X) \approx 1.935$
• (b) $P(X \leq 4) \approx 0.831$
• (c) $P(2.5 \leq X \leq 4) \approx 0.331$

Would you like further details on any of these steps?

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Related Questions:

1. How does changing $\alpha$ affect the shape of the Weibull distribution?
2. How would $E(X)$ and $V(X)$ change if $\beta$ were different?
3. How is the Weibull distribution used in reliability analysis?
4. What would $P(X > 4)$ be in this scenario?
5. Can the Gamma function be approximated differently for better accuracy?

Tip:

The Weibull distribution is often used for modeling lifetimes, where different values of $\alpha$ and $\beta$ can indicate various failure rates.