1. **Stating the problem:** We have monthly rainfalls modeled as independent and identically distributed normal random variables with mean $\mu = 20$ cm and variance $\sigma^2 = 12$ cm$^2$. We want to find: - The probability that the total rainfall over 16 months is 220 cm. - The probability that the rainfall in a single month is less than 18 cm. 2. **Relevant formulas and rules:** - The sum of $n$ independent normal variables $X_i \sim N(\mu, \sigma^2)$ is also normal: $S_n = \sum_{i=1}^n X_i \sim N(n\mu, n\sigma^2)$. - For a normal variable $X \sim N(\mu, \sigma^2)$, the standardized variable is $Z = \frac{X - \mu}{\sigma} \sim N(0,1)$. - Probability of an exact value in a continuous distribution is 0, so we interpret the first question as probability of total rainfall being less than or equal to 220 cm. 3. **Distribution of total rainfall over 16 months:** - Mean: $$\mu_{16} = 16 \times 20 = 320$$ cm - Variance: $$\sigma^2_{16} = 16 \times 12 = 192$$ cm$^2$ - Standard deviation: $$\sigma_{16} = \sqrt{192} = 13.8564$$ cm 4. **Probability total rainfall $\leq$ 220 cm:** - Standardize total rainfall: $$Z = \frac{220 - 320}{13.8564} = \frac{-100}{13.8564} = -7.22$$ - Using standard normal distribution tables or calculator, $$P(Z \leq -7.22) \approx 0$$ (extremely small probability). 5. **Probability monthly rainfall $<$ 18 cm:** - For one month, $X \sim N(20, 12)$, standard deviation $$\sigma = \sqrt{12} = 3.4641$$ cm - Standardize: $$Z = \frac{18 - 20}{3.4641} = \frac{-2}{3.4641} = -0.577$$ - From standard normal tables, $$P(Z < -0.577) \approx 0.281$$. **Final answers:** - Probability total rainfall over 16 months is 220 cm or less is approximately 0. - Probability monthly rainfall is less than 18 cm is approximately 0.281.