Problem statements: A pharmaceutical company tested a new vaccine on 20 children and 15 developed immunity; additional questions concern birth weights (normal), sample means (SBP), and length of stay (LOS). 1. Vaccine part a — State the problem and model. Assume each child independently develops immunity with probability $p=15/20=0.75$ estimated from the sample, so $X\sim\mathrm{Bin}(n=20,p=0.75)$ and we seek $P(X\ge 15)$. The probability is given by the binomial sum. $$P(X\ge 15)=\sum_{k=15}^{20} \binom{20}{k}0.75^{k}0.25^{20-k}$$ Compute each term approximately. $$\begin{aligned}P(X=15)&=\binom{20}{15}0.75^{15}0.25^{5}\approx 0.2025,\\P(X=16)&\approx 0.1899,\\P(X=17)&\approx 0.1340,\\P(X=18)&\approx 0.0670,\\P(X=19)&\approx 0.0212,\\P(X=20)&\approx 0.0032.\end{aligned}$$ Sum the terms to get the final probability. $$P(X\ge 15)\approx 0.2025+0.1899+0.1340+0.0670+0.0212+0.0032\approx 0.6177.$$ Final answer for 1a: $P(X\ge 15)\approx 0.6177$. 2. Vaccine part b — Expected value and standard deviation. For a binomial $X\sim\mathrm{Bin}(n,p)$ we have $E[X]=np$ and $\mathrm{Var}(X)=np(1-p)$. Compute with $n=20$ and $p=0.75$. $$E[X]=20\times 0.75=15.$$ $$\mathrm{SD}(X)=\sqrt{\mathrm{Var}(X)}=\sqrt{20\times 0.75\times 0.25}=\sqrt{3.75}\approx 1.9365.$$ Final answers for 1b: expected number $=15$ and standard deviation $\approx 1.9365$. 3. Birth weights part a — State the problem and method. Birth weights are $\mathrm{Normal}(\mu=3.1,\sigma=0.5)$ and we want the 10th percentile $x_{0.10}$ such that $P(X\le x_{0.10})=0.10$. Use the standard normal quantile $z_{0.10}\approx -1.28155$ and transform. $$x_{0.10}=\mu+z_{0.10}\sigma=3.1+(-1.28155)\times 0.5\approx 3.1-0.64078\approx 2.4592\,\text{kg}.$$ Final answer for 2a: $x_{0.10}\approx 2.46\,$kg. 4. Birth weights part b — Expected count under 2.5 kg out of 400. Compute $P(X<2.5)$ where $Z=(2.5-3.1)/0.5=-1.2$. $$P(X<2.5)=\Phi(-1.2)\approx 0.11507.$$ Expected number out of 400 births is $400\times 0.11507\approx 46.03$, so about 46 babies. Final answer for 2b: approximately 46 underweight babies per 400 births. 5. SBP part a — Sampling distribution mean and standard error. For population mean $\mu=130$ and population SD $\sigma=15$ with sample size $n=40$, the sampling distribution of the sample mean $\bar{X}$ has mean $\mu_{\bar{X}}=\mu$ and standard error $\sigma_{\bar{X}}=\sigma/\sqrt{n}$. $$\mu_{\bar{X}}=130.$$ $$\sigma_{\bar{X}}=\frac{15}{\sqrt{40}}\approx \frac{15}{6.3246}\approx 2.3717.$$ Final answers for 3a: mean $=130$ and standard error $\approx 2.3717$. 6. SBP part b — Probability sample mean exceeds 135. Compute $P(\bar{X}>135)$ using $Z=(\bar{X}-\mu)/\sigma_{\bar{X}}$. $$Z=\frac{135-130}{2.3717}\approx 2.1082.$$ Upper-tail probability $P(Z>2.1082)\approx 0.0175$. Final answer for 3b: $P(\bar{X}>135)\approx 0.0175$ (about 1.75%). 7. LOS part a — 95% probability interval for the sample mean using the CLT. Even if LOS is skewed, the CLT implies the sampling distribution of $\bar{X}$ is approximately normal for moderate/large $n$ with mean $\mu=5$ and standard error $\sigma/\sqrt{n}=3/\sqrt{n}$. A 95% probability (confidence) interval for $\bar{X}$ is approximately $$\mu\pm 1.96\frac{\sigma}{\sqrt{n}}=5\pm 1.96\frac{3}{\sqrt{n}}.$$ If a concrete sample size is needed, for example $n=30$, then $$\text{SE}=\frac{3}{\sqrt{30}}\approx 0.5477,\quad \text{margin}=1.96\times 0.5477\approx 1.0725,$$ so the 95% interval is approximately $(5-1.0725,\;5+1.0725)=(3.9275,6.0725)$. Final answer for 4a: general interval $5\pm 1.96\frac{3}{\sqrt{n}}$; for $n=30$ approximately $(3.93,6.07)$. 8. LOS part b — Real-world implications of the CLT for hospital management. The CLT means that averages of LOS from reasonably large random samples behave approximately normally even when individual stays are skewed. This allows managers to predict variability in average LOS with the standard error $3/\sqrt{n}$ and to compute probabilities or intervals for average demand. In practice, knowing the distribution of average LOS helps schedule staff and allocate beds because planners can estimate the likely range of average occupancy and prepare for typical fluctuations rather than extreme individual cases. For example, increasing the sample window (larger $n$) reduces the SE and yields more precise estimates of average LOS, which supports more reliable staffing and resource decisions. Final answer for 4b: use CLT-based intervals and SE calculations to inform staffing, bed allocation, and surge planning based on expected variability.