1. **Problem Statement:** We have storage devices with capacities following a normal distribution with mean $\mu = 12$ terabytes and standard deviation $\sigma = 1.5$ terabytes. 2. **Define the random variable:** Let $X$ be the random variable representing the capacity of a storage device in terabytes. Then, $X \sim N(12, 1.5^2)$. 3. **Probability calculations use the standard normal distribution:** We convert $X$ to the standard normal variable $Z$ using: $$Z = \frac{X - \mu}{\sigma}$$ 4. **(f) Probability that capacity is more than 15 terabytes:** Calculate $P(X > 15)$. Calculate the $Z$-score: $$Z = \frac{15 - 12}{1.5} = \frac{3}{1.5} = 2$$ Using standard normal tables or a calculator: $$P(Z > 2) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228$$ So, $P(X > 15) = 0.0228$. 5. **(g) Probability that capacity is less than 7.5 terabytes:** Calculate $P(X < 7.5)$. Calculate the $Z$-score: $$Z = \frac{7.5 - 12}{1.5} = \frac{-4.5}{1.5} = -3$$ Using standard normal tables: $$P(Z < -3) = 0.0013$$ So, $P(X < 7.5) = 0.0013$. 6. **(h) Probability that capacity is between 9 and 15 terabytes:** Calculate $P(9 < X < 15)$. Calculate $Z$-scores: $$Z_1 = \frac{9 - 12}{1.5} = -2$$ $$Z_2 = \frac{15 - 12}{1.5} = 2$$ Using standard normal tables: $$P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) = 0.9772 - 0.0228 = 0.9544$$ So, $P(9 < X < 15) = 0.9544$. **Final answers:** - $P(X > 15) = 0.0228$ - $P(X < 7.5) = 0.0013$ - $P(9 < X < 15) = 0.9544$