1. Problem: A vaccine trial with 20 children, 15 developed immunity. (a) Probability at least 15 develop immunity. (b) Expected number and standard deviation.\n 1. (a) Let $X \sim \text{Binomial}(n=20, p)$ with unknown $p$, assume sample proportion $\hat p=15/20=0.75$. Probability exactly $k$ successes: $$P(X=k) = \binom{20}{k}p^k(1-p)^{20-k}.$$ For "at least 15": $$P(X \geq 15) = \sum_{k=15}^{20} P(X=k).$$ Without $p$, assume $p=0.75$, compute numeric sum (approximation): use binomial cumulative functions or software.\n (b) Expected number: $$E(X) = np = 20 \times 0.75 = 15.$$\n Standard deviation: $$\sigma = \sqrt{np(1-p)} = \sqrt{20 \times 0.75 \times 0.25} = \sqrt{3.75} \approx 1.936.$$\n 2. Normal distribution of birth weights: mean $\mu=3.1$, SD $\sigma=0.5$.\n (a) Find weight delimiting lightest 10%. We seek $x$ with $$P(X \leq x) = 0.10,$$ thus $x = \mu + z_{0.10} \sigma,$ where $z_{0.10} \approx -1.281$.\n$$x = 3.1 + (-1.281)(0.5) = 3.1 - 0.6405 = 2.4595 \text{ kg}.$$\n (b) "Underweight" means under 2.5 kg. Find $P(X < 2.5)$:\n$$z = \frac{2.5 - 3.1}{0.5} = -1.2,$$ so $P = \Phi(-1.2) \approx 0.1151.$\nFor 400 births, expected underweight babies: $$400 \times 0.1151 = 46.04 \approx 46.$$\n 3. Sampling distribution of mean SBP, $n=40$, population mean $130$, SD $15$.\n (a) Mean of sample mean: $$\mu_{\bar{X}} = 130.$$\nStandard error: $$\text{SE} = \frac{15}{\sqrt{40}} \approx 2.371.$$\n (b) Probability sample mean $> 135$:\nCalculate z-score:\n$$z = \frac{135 -130}{2.371} \approx 2.11.$$\nLooking up standard normal, \(P(Z>2.11)\approx 0.0174.$\n 4. Length of stay (LOS): mean $5$ days, SD $3$ days, right-skewed, sample mean behavior.\n (a) 95% probability interval for sample mean with sample size $n$ (not given, assume large):\nUsing Central Limit Theorem (CLT), if sample size $n$ is large, sample mean $\bar{X} \sim N(5, \frac{3}{\sqrt{n}})$. Since $n$ not given, assuming $n=30$ for example:\nStandard error: $$SE = \frac{3}{\sqrt{30}} = 0.5477.$$\n95% CI: $$5 \pm 1.96 \times 0.5477 = (3.93, 6.07).$$\n (b) The CLT implies sample means approximate normality even if individual LOS is skewed, enabling hospital management to predict average LOS reliably. This informs staffing and resource planning by understanding likely demand and avoiding under- or overstaffing situations.