1. Problem Statement: We have two normal distributions: - $X \sim \mathcal{N}(44, 2^2)$ means $X$ is normally distributed with mean $\mu=44$ and standard deviation $\sigma=2$. - $X \sim \mathcal{N}(66, 11^2)$ means $X$ is normally distributed with mean $\mu=66$ and standard deviation $\sigma=11$. We need to calculate probabilities from these distributions rounded to 3 decimal places. --- ### First distribution: $X \sim \mathcal{N}(44,4)$ 2. Calculate $P(X \leq 47.2)$: - Compute z-score: $z = \frac{47.2 - 44}{2} = \frac{3.2}{2} = 1.6$ - From standard normal tables or calculator, $P(Z \leq 1.6) = 0.9452$ - So, $P(X \leq 47.2) = 0.945$ 3. Calculate $P(X > 44.8)$: - $z = \frac{44.8 - 44}{2} = 0.4$ - $P(Z > 0.4) = 1 - P(Z \leq 0.4) = 1 - 0.6554 = 0.3446$ - So, $P(X > 44.8) = 0.345$ 4. Calculate $P(X > 41.6)$: - $z = \frac{41.6 - 44}{2} = -1.2$ - $P(Z > -1.2) = 1 - P(Z \leq -1.2) = 1 - 0.1151 = 0.8849$ - So, $P(X > 41.6) = 0.885$ 5. Calculate $P(X < 39)$: - $z = \frac{39 - 44}{2} = -2.5$ - $P(Z < -2.5) = 0.0062$ - So, $P(X < 39) = 0.006$ --- ### Second distribution: $X \sim \mathcal{N}(66,11^2)$ 6. Calculate $P(X \leq 88)$: - $z = \frac{88 - 66}{11} = 2$ - $P(Z \leq 2) = 0.9772$ - So, $P(X \leq 88) = 0.977$ 7. Calculate $P(X \leq 59.4)$: - $z = \frac{59.4 - 66}{11} = -0.6$ - $P(Z \leq -0.6) = 0.2743$ - So, $P(X \leq 59.4) = 0.274$ 8. Calculate $P(X > 51.7)$: - $z = \frac{51.7 - 66}{11} = -1.3$ - $P(Z > -1.3) = 1 - P(Z \leq -1.3) = 1 - 0.0968 = 0.9032$ - So, $P(X > 51.7) = 0.903$ 9. Calculate $P(64.9 < X < 73.7)$: - $z_1 = \frac{64.9 - 66}{11} = -0.1$ - $z_2 = \frac{73.7 - 66}{11} = 0.7$ - $P(-0.1 < Z < 0.7) = P(Z \leq 0.7) - P(Z \leq -0.1) = 0.7580 - 0.4602 = 0.2978$ - So, $P(64.9 < X < 73.7) = 0.298$ Final Answers: - $P(X \leq 47.2) = 0.945$ - $P(X > 44.8) = 0.345$ - $P(X > 41.6) = 0.885$ - $P(X < 39) = 0.006$ - $P(X \leq 88) = 0.977$ - $P(X \leq 59.4) = 0.274$ - $P(X > 51.7) = 0.903$ - $P(64.9 < X < 73.7) = 0.298$