1. **Problem 1: Heights of adult males** Given: Heights are normally distributed with mean $\mu=180$ cm and standard deviation $\sigma$ cm (unknown). - 17% are shorter than 168 cm. - 80% have heights between $(192 - h)$ cm and 192 cm. Find $h$. 2. **Step 1: Use the first condition to find $\sigma$** The probability that height $X < 168$ cm is 0.17. Standardize: $$P\left(Z < \frac{168 - 180}{\sigma}\right) = 0.17$$ From standard normal tables, $P(Z < z) = 0.17$ corresponds to $z \approx -0.95$. So, $$\frac{168 - 180}{\sigma} = -0.95$$ Simplify: $$\frac{-12}{\sigma} = -0.95 \implies \sigma = \frac{12}{0.95} \approx 12.63$$ 3. **Step 2: Use the second condition to find $h$** We know 80% of men have heights between $(192 - h)$ and 192 cm. Standardize these bounds: Upper bound: $$z_2 = \frac{192 - 180}{12.63} = \frac{12}{12.63} \approx 0.95$$ Lower bound: $$z_1 = \frac{192 - h - 180}{12.63} = \frac{12 - h}{12.63}$$ The probability between these bounds is 0.80: $$P(z_1 < Z < z_2) = 0.80$$ From standard normal tables, $$P(Z < 0.95) \approx 0.83$$ So, $$P(Z < z_1) = 0.83 - 0.80 = 0.03$$ From tables, $P(Z < z_1) = 0.03$ corresponds to $z_1 \approx -1.88$. Set equal: $$\frac{12 - h}{12.63} = -1.88$$ Multiply both sides: $$12 - h = -1.88 \times 12.63$$ $$12 - h = -23.74$$ Solve for $h$: $$h = 12 + 23.74 = 35.74$$ \boxed{h \approx 35.74 \text{ cm}}$ --- 4. **Problem 2: Weights of oranges** Given: Weights $X$ are normally distributed with mean $\mu=297$ g. - 79% weigh more than 289 g. - 9.5% weigh more than 310 g. (a) Find $P(289 < X < 310)$. 5. **Step 1: Find standard deviation $\sigma$** From $P(X > 289) = 0.79$, we get $P(X \leq 289) = 0.21$. Standardize: $$P\left(Z \leq \frac{289 - 297}{\sigma}\right) = 0.21$$ From tables, $P(Z \leq z) = 0.21$ corresponds to $z \approx -0.81$. So, $$\frac{289 - 297}{\sigma} = -0.81 \implies \frac{-8}{\sigma} = -0.81 \implies \sigma = \frac{8}{0.81} \approx 9.88$$ 6. **Step 2: Find $P(289 < X < 310)$** Standardize bounds: Lower bound: $$z_1 = \frac{289 - 297}{9.88} = \frac{-8}{9.88} \approx -0.81$$ Upper bound: $$z_2 = \frac{310 - 297}{9.88} = \frac{13}{9.88} \approx 1.32$$ Find probabilities from tables: $$P(Z < z_2) \approx P(Z < 1.32) = 0.9066$$ $$P(Z < z_1) = 0.21$$ So, $$P(289 < X < 310) = P(z_1 < Z < z_2) = 0.9066 - 0.21 = 0.6966$$ \boxed{P(289 < X < 310) \approx 0.697}$ --- **Final answers:** - $h \approx 35.74$ cm - $P(289 < X < 310) \approx 0.697$