1. **Problem statement:** We have a normal distribution of reading times with mean $\mu = 40$ minutes and variance $\sigma^2 = 36$ minutes², so standard deviation $\sigma = \sqrt{36} = 6$ minutes. We want to find probabilities for: a. $P(X < 32)$ b. $P(42 < X < 51)$ c. $P(X > 36)$ 2. **Formula and rules:** The standard normal variable $Z$ is defined as: $$Z = \frac{X - \mu}{\sigma}$$ We use the cumulative distribution function (CDF) of the standard normal distribution to find probabilities. The table given shows $P(0 < Z < z)$ for positive $z$. Important rules: - For $Z < 0$, use symmetry: $P(Z < -z) = 0.5 - P(0 < Z < z)$ - For $Z > 0$, $P(Z < z) = 0.5 + P(0 < Z < z)$ - Total area under the curve is 1. 3. **Calculations:** **a. Find $P(X < 32)$:** Calculate $Z$: $$Z = \frac{32 - 40}{6} = \frac{-8}{6} = -1.33$$ From the table, find $P(0 < Z < 1.33)$: - Look at row 1.3 and column 0.03: value is approximately 0.4082 Since $Z$ is negative: $$P(Z < -1.33) = 0.5 - 0.4082 = 0.0918$$ So, $$P(X < 32) = 0.0918$$ **b. Find $P(42 < X < 51)$:** Calculate $Z$ for 42: $$Z_1 = \frac{42 - 40}{6} = \frac{2}{6} = 0.33$$ Calculate $Z$ for 51: $$Z_2 = \frac{51 - 40}{6} = \frac{11}{6} \approx 1.83$$ From the table: - $P(0 < Z < 0.33)$: row 0.3, col 0.03 = 0.1293 - $P(0 < Z < 1.83)$: row 1.8, col 0.03 = 0.4664 Probability between 42 and 51: $$P(0.33 < Z < 1.83) = P(0 < Z < 1.83) - P(0 < Z < 0.33) = 0.4664 - 0.1293 = 0.3371$$ **c. Find $P(X > 36)$:** Calculate $Z$: $$Z = \frac{36 - 40}{6} = \frac{-4}{6} = -0.67$$ From the table: - $P(0 < Z < 0.67)$: row 0.6, col 0.07 = 0.2517 Since $Z$ is negative: $$P(Z < -0.67) = 0.5 - 0.2517 = 0.2483$$ Therefore, $$P(Z > -0.67) = 1 - P(Z < -0.67) = 1 - 0.2483 = 0.7517$$ So, $$P(X > 36) = 0.7517$$ **Final answers:** - a. $0.0918$ - b. $0.3371$ - c. $0.7517$