1. **Problem Statement:** A) Find the 14th percentile (P14) for the standard normal distribution $Z$. B) Find the 95th percentile (P95) for the $t$-distribution with 12 degrees of freedom ($t_{12}$). 2. **Formulas and Important Rules:** - The percentile $P_p$ is the value below which $p\%$ of the data falls. - For the standard normal distribution $Z$, percentiles correspond to $z$-scores found using the inverse cumulative distribution function (CDF). - For the $t$-distribution with $df$ degrees of freedom, percentiles correspond to $t$-scores found using the inverse CDF of the $t$-distribution. 3. **Step A: Find $P14$ for $Z$** - We want $z$ such that $P(Z \leq z) = 0.14$. - Using standard normal tables or a calculator, find $z = \Phi^{-1}(0.14)$. - From tables or software, $z \approx -1.080$ (rounded to 3 decimals). 4. **Step B: Find $P95$ for $t_{12}$** - We want $t$ such that $P(T \leq t) = 0.95$ with $df=12$. - Using $t$-distribution tables or software, find $t = t_{0.95,12}$. - From tables or software, $t \approx 1.782$ (rounded to 3 decimals). **Final answers:** - $P14$ for $Z$ is approximately **$-1.080$**. - $P95$ for $t_{12}$ is approximately **$1.782$**. These values represent the cutoff points below which 14% and 95% of the respective distributions fall.