1. **Define the continuous random variable:** Let $X$ be the length of a Taylor Swift song in seconds. Since song lengths vary continuously, $X$ is a continuous random variable. 2. **Find $P(X \leq 160)$:** a) Probability statement: $P(X \leq 160)$. b) This corresponds to the area under the normal distribution curve to the left of 160 seconds. c) To find this probability, we calculate the z-score: $$z = \frac{160 - \mu}{\sigma}$$ Assuming the mean $\mu = 200$ seconds and standard deviation $\sigma = 30$ seconds (typical values for song lengths), $$z = \frac{160 - 200}{30} = \frac{-40}{30} = -1.33$$ Using the z-score table, $P(Z \leq -1.33) \approx 0.0918$. 3. **Find $P(180 \leq X \leq 220)$:** a) Probability statement: $P(180 \leq X \leq 220)$. b) This is the area under the curve between 180 and 220 seconds. c) Calculate z-scores: $$z_1 = \frac{180 - 200}{30} = -0.67$$ $$z_2 = \frac{220 - 200}{30} = 0.67$$ From the z-table: $P(Z \leq 0.67) \approx 0.7486$ $P(Z \leq -0.67) \approx 0.2514$ So, $$P(180 \leq X \leq 220) = 0.7486 - 0.2514 = 0.4972$$ 4. **Find $P(X > 226)$ where 226 seconds = 3 minutes 46 seconds:** a) Probability statement: $P(X > 226)$. b) This is the area under the curve to the right of 226 seconds. c) Calculate z-score: $$z = \frac{226 - 200}{30} = \frac{26}{30} = 0.87$$ From the z-table: $P(Z \leq 0.87) \approx 0.8078$ Therefore, $$P(X > 226) = 1 - 0.8078 = 0.1922$$ **Final answers:** - $P(X \leq 160) \approx 0.0918$ - $P(180 \leq X \leq 220) \approx 0.4972$ - $P(X > 226) \approx 0.1922$