1. Problem: Find $z$ such that $P(X > z + \mu_x) = \frac{1}{4}$ for $X \sim \text{Uniform}(1,2)$.\nStep 1: Mean of uniform $\mu_x = \frac{1+2}{2} = 1.5$.\nStep 2: We want $P(X > z + 1.5) = \frac{1}{4}$.\nStep 3: For uniform, $P(X > a) = \frac{2 - a}{2 - 1} = 2 - a$.\nStep 4: Set $2 - (z + 1.5) = \frac{1}{4} \Rightarrow z + 1.5 = 1.75 \Rightarrow z = 0.25$.\n\n2. Problem: Find $P(|X - 2| < 1)$ for $X \sim N(2,1)$.\nStep 1: $P(|X - 2| < 1) = P(1 < X < 3)$.\nStep 2: Standardize: $Z = \frac{X - 2}{1}$.\nStep 3: $P(1 < X < 3) = P(-1 < Z < 1)$.\nStep 4: Using standard normal table, $P(-1 < Z < 1) = 0.6826$.\n\n3. Problem: Find $P(X \leq \mu_x - 3\sigma_x)$ for $X \sim \text{Binomial}(100,0.1)$.\nStep 1: Mean $\mu_x = np = 10$, variance $\sigma_x^2 = np(1-p) = 9$, so $\sigma_x = 3$.\nStep 2: $\mu_x - 3\sigma_x = 10 - 9 = 1$.\nStep 3: $P(X \leq 1) = P(X=0) + P(X=1)$.\nStep 4: $P(X=0) = (0.9)^{100}$ (very small), $P(X=1) = 100 \times 0.1 \times (0.9)^{99}$.\nStep 5: Sum these for final probability (approx $\approx 0.00004$).\n\n4. Problem: If $P(X=0) = \frac{1}{2}$ for Poisson $X$, find $E(X)$.\nStep 1: $P(X=0) = e^{-\lambda} = \frac{1}{2}$.\nStep 2: Solve $e^{-\lambda} = 0.5 \Rightarrow \lambda = \ln 2 \approx 0.693$.\nStep 3: $E(X) = \lambda = 0.693$.\n\n5. Problem: For binomial $X \sim Bin(n,p)$, find $p$ maximizing $Var(X)$.\nStep 1: $Var(X) = np(1-p)$.\nStep 2: For fixed $n$, maximize $p(1-p)$.\nStep 3: Derivative $1 - 2p = 0 \Rightarrow p = 0.5$.\n\n6. Problem: Binomial with mean 6 and std dev $\sqrt{2}$. Find first two terms $P(X=0)$ and $P(X=1)$.\nStep 1: Mean $= np = 6$, variance $= np(1-p) = 2$.\nStep 2: From mean and variance, $p = 1 - \frac{2}{6} = \frac{2}{3}$, $n = \frac{6}{p} = 9$.\nStep 3: $P(X=0) = (1-p)^n = (1/3)^9$.\nStep 4: $P(X=1) = n p (1-p)^{n-1} = 9 \times \frac{2}{3} \times (1/3)^8$.\n\n7. Problem: For Poisson $X$, $P(X=2) = \frac{2}{3} P(X=1)$, find $P(X=3)$.\nStep 1: $P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!}$.\nStep 2: $P(2) = \frac{2}{3} P(1) \Rightarrow e^{-\lambda} \frac{\lambda^2}{2} = \frac{2}{3} e^{-\lambda} \lambda$.\nStep 3: Simplify: $\frac{\lambda^2}{2} = \frac{2}{3} \lambda \Rightarrow \lambda = \frac{4}{3}$.\nStep 4: $P(3) = e^{-\lambda} \frac{\lambda^3}{3!} = e^{-4/3} \frac{(4/3)^3}{6}$.\n\n8. Problem: $X \sim N(200, \sigma^2)$, find largest $\sigma$ so that $P(X > 150) \geq 0.9$.\nStep 1: $P(X > 150) = 0.9 \Rightarrow P(X \leq 150) = 0.1$.\nStep 2: Standardize: $Z = \frac{150 - 200}{\sigma} = -\frac{50}{\sigma}$.\nStep 3: $P(Z \leq -\frac{50}{\sigma}) = 0.1$.\nStep 4: From Z-table, $z_{0.1} = -1.2816$.\nStep 5: $-\frac{50}{\sigma} = -1.2816 \Rightarrow \sigma = \frac{50}{1.2816} \approx 39.01$.\n\n9. Problem: Poisson with mean 1 accident/day, find:\na) $P(>10)$ accidents in a week.\nb) $P(>2)$ days lapse between accidents.\nStep 1a: Weekly mean $\lambda = 7$.\nStep 2a: $P(X > 10) = 1 - P(X \leq 10)$, use Poisson CDF with $\lambda=7$.\nStep 3a: Approximate $P(X \leq 10) \approx 0.89$, so $P(X > 10) \approx 0.11$.\nStep 1b: Days between accidents is exponential with rate 1.\nStep 2b: $P(>2) = e^{-2} \approx 0.1353$.