1. **State the problem:** We have plums with diameters normally distributed. We know 60% have diameter less than 5.9 cm and 20% have diameter greater than 6.1 cm. We need to find the mean $\mu$ and standard deviation $\sigma$ of this distribution. 2. **Recall the normal distribution properties:** For a normal distribution $X \sim N(\mu, \sigma^2)$, the standardized variable is $Z = \frac{X - \mu}{\sigma}$ which follows $N(0,1)$. 3. **Translate given probabilities to $Z$-scores:** - $P(X < 5.9) = 0.60$ means $P\left(Z < \frac{5.9 - \mu}{\sigma}\right) = 0.60$. - $P(X > 6.1) = 0.20$ means $P\left(Z > \frac{6.1 - \mu}{\sigma}\right) = 0.20$ or equivalently $P\left(Z < \frac{6.1 - \mu}{\sigma}\right) = 0.80$. 4. **Find $Z$-values from standard normal table or inverse CDF:** - $z_1 = \Phi^{-1}(0.60) \approx 0.253$ - $z_2 = \Phi^{-1}(0.80) \approx 0.842$ 5. **Set up equations:** $$\frac{5.9 - \mu}{\sigma} = 0.253$$ $$\frac{6.1 - \mu}{\sigma} = 0.842$$ 6. **Solve the system:** Subtract first from second: $$\frac{6.1 - \mu}{\sigma} - \frac{5.9 - \mu}{\sigma} = 0.842 - 0.253$$ $$\frac{6.1 - 5.9}{\sigma} = 0.589$$ $$\frac{0.2}{\sigma} = 0.589$$ $$\sigma = \frac{0.2}{0.589} \approx 0.34$$ 7. **Find $\mu$ using first equation:** $$5.9 - \mu = 0.253 \times 0.34$$ $$5.9 - \mu = 0.086$$ $$\mu = 5.9 - 0.086 = 5.814$$ **Final answer:** $$\mu \approx 5.81 \text{ cm}, \quad \sigma \approx 0.34 \text{ cm}$$