Repair Time Probability 8C7349

1. **Problem Statement:** We have two phases of repair for each machine: - Phase 1 time $X_1$ is exponentially distributed with mean $0.3$ hours. - Phase 2 time $X_2$ is exponentially distributed with mean $0.1$ hours. The total repair time for one machine is $X = X_1 + X_2$. We want the probability that the worker repairs $20$ machines within $10$ hours. 2. **Key facts and formulas:** - If $X_i \sim \text{Exponential}(\lambda_i)$, then mean $= \frac{1}{\lambda_i}$. - For phase 1: mean $0.3$ implies $\lambda_1 = \frac{1}{0.3} = \frac{10}{3}$. - For phase 2: mean $0.1$ implies $\lambda_2 = \frac{1}{0.1} = 10$. - The sum of two independent exponential random variables with different rates follows a hypoexponential distribution. 3. **Distribution of total time for one machine:** $X = X_1 + X_2$ where $X_1 \sim \text{Exp}(\lambda_1)$ and $X_2 \sim \text{Exp}(\lambda_2)$ independent. 4. **Sum of 20 machines:** Let $T = \sum_{i=1}^{20} X_i$ be the total time to repair 20 machines. Since machines are independent and identically distributed, $T$ is the sum of 20 i.i.d. hypoexponential random variables. 5. **Approximating the distribution of $X$:** The hypoexponential distribution is complicated, so we use the mean and variance to approximate $X$ by a Gamma distribution. - Mean of $X$: $$E[X] = E[X_1] + E[X_2] = 0.3 + 0.1 = 0.4$$ - Variance of $X$: $$\text{Var}(X) = \text{Var}(X_1) + \text{Var}(X_2) = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} = 0.3^2 + 0.1^2 = 0.09 + 0.01 = 0.1$$ 6. **Sum of 20 such variables:** - Mean of $T$: $$E[T] = 20 \times 0.4 = 8$$ - Variance of $T$: $$\text{Var}(T) = 20 \times 0.1 = 2$$ 7. **Gamma approximation for $T$:** The Gamma distribution has parameters shape $k$ and scale $\theta$ with: $$E[T] = k\theta, \quad \text{Var}(T) = k\theta^2$$ From these: $$k = \frac{(E[T])^2}{\text{Var}(T)} = \frac{8^2}{2} = \frac{64}{2} = 32$$ $$\theta = \frac{\text{Var}(T)}{E[T]} = \frac{2}{8} = 0.25$$ 8. **Calculate the probability:** We want: $$P(T \leq 10) = F_{\text{Gamma}}(10; k=32, \theta=0.25)$$ Using the Gamma CDF: $$P(T \leq 10) = \text{GammaCDF}(10; 32, 0.25)$$ 9. **Interpretation:** Since $E[T] = 8$ and $10$ is greater than the mean, the probability will be reasonably high. **Final answer:** $$\boxed{P(T \leq 10) = \text{GammaCDF}(10; 32, 0.25)}$$ This can be computed using statistical software or tables for the Gamma distribution.