1. **Fish length for restaurants:** The fish lengths are normally distributed with mean $\mu=48$ cm and standard deviation $\sigma=5$ cm. We want the longest 36% of fish, so we find the cutoff length $x$ such that $P(X>x)=0.36$ or $P(X\leq x)=0.64$. Using the standard normal distribution, find $z$ where $P(Z\leq z)=0.64$. From z-tables or calculator, $z \approx 0.36$. Use the formula $x=\mu + z\sigma = 48 + 0.36 \times 5 = 48 + 1.8 = 49.8$ cm. So fish longer than **49.8 cm** go to restaurants. 2. **Timber length setting:** We want 91% of pieces to be longer than 780 mm, so $P(X>780)=0.91$ or $P(X\leq 780)=0.09$. Find $z$ such that $P(Z\leq z)=0.09$. From z-tables, $z \approx -1.34$. Use $x=\mu + z\sigma$, where $x=780$, $\sigma=4$ mm. Rearranged: $\mu = x - z\sigma = 780 - (-1.34) \times 4 = 780 + 5.36 = 785.36$ mm. Rounded to nearest whole number: **785 mm**. **Final answers:** - Fish length cutoff: **49.8 cm** - Timber mean setting: **785 mm**