1. **State the problem:** We have weights of oranges normally distributed with mean $\mu=297$ g. Given that 79% weigh more than 289 g and 9.5% weigh more than 310 g, find the probability an orange weighs between 289 g and 310 g. 2. **Identify known values and notation:** Let $X$ be the weight of an orange, $X \sim N(\mu, \sigma^2)$ with mean $\mu=297$ g and unknown standard deviation $\sigma$. We know: - $P(X > 289) = 0.79$ - $P(X > 310) = 0.095$ 3. **Convert to standard normal variable $Z$:** $$Z = \frac{X - \mu}{\sigma}$$ Then: $$P(X > 289) = P\left(Z > \frac{289 - 297}{\sigma}\right) = 0.79$$ $$P(X > 310) = P\left(Z > \frac{310 - 297}{\sigma}\right) = 0.095$$ 4. **Find corresponding $Z$-scores from standard normal table:** - Since $P(Z > z_1) = 0.79$, then $P(Z \leq z_1) = 1 - 0.79 = 0.21$. From the standard normal table, $z_1 \approx -0.81$ (because 0.21 is left tail). - Since $P(Z > z_2) = 0.095$, then $P(Z \leq z_2) = 1 - 0.095 = 0.905$. From the table, $z_2 \approx 1.31$. 5. **Set up equations for $\sigma$:** $$\frac{289 - 297}{\sigma} = z_1 = -0.81$$ $$\frac{310 - 297}{\sigma} = z_2 = 1.31$$ 6. **Solve for $\sigma$ from each equation:** $$\sigma = \frac{289 - 297}{-0.81} = \frac{-8}{-0.81} = 9.88$$ $$\sigma = \frac{310 - 297}{1.31} = \frac{13}{1.31} = 9.92$$ 7. **Average $\sigma$ values for accuracy:** $$\sigma \approx \frac{9.88 + 9.92}{2} = 9.90$$ 8. **Find the probability that $X$ is between 289 g and 310 g:** $$P(289 < X < 310) = P\left(\frac{289 - 297}{9.90} < Z < \frac{310 - 297}{9.90}\right) = P(-0.81 < Z < 1.31)$$ 9. **Calculate this probability using standard normal CDF $\Phi$:** $$P(-0.81 < Z < 1.31) = \Phi(1.31) - \Phi(-0.81)$$ From tables: $$\Phi(1.31) = 0.905$$ $$\Phi(-0.81) = 1 - \Phi(0.81) = 1 - 0.791 = 0.209$$ 10. **Final probability:** $$P(289 < X < 310) = 0.905 - 0.209 = 0.696$$ \boxed{\text{The probability that an orange weighs between 289 g and 310 g is approximately } 0.696}