1. **Stating the problem:** Andy believes 72% of his customers are satisfied. From a sample of 92, we want probabilities of satisfaction numbers. Let $X$ = number of satisfied customers, $X \sim \text{Binomial}(n=92, p=0.72)$. 2. **Approximate Binomial with Normal distribution:** Mean $\mu = np = 92 \times 0.72 = 66.24$. Standard deviation $\sigma = \sqrt{np(1-p)} = \sqrt{92 \times 0.72 \times 0.28} = \sqrt{18.5472} \approx 4.3064$. 3. Use Normal approximation with continuity correction. (a) Probability $P(X < 67)$ = $P(X \leq 66)$ with correction, evaluate $P(Y \leq 66.5)$ where $Y$ is normal. Calculate Z-score: $$Z = \frac{66.5 - 66.24}{4.3064} = \frac{0.26}{4.3064} \approx 0.0604$$ Using standard normal table, $P(Z \leq 0.0604) \approx 0.5241$. (b) Probability $P(X \geq 67)$ = $1 - P(X \leq 66)$, so $$P(X \geq 67) = 1 - 0.5241 = 0.4759$$ (c) Sample proportion range 70% to 77%: Convert to number satisfied: Lower bound $= 0.70 \times 92 = 64.4$, upper bound $= 0.77 \times 92 = 70.84$. Using continuity correction: Find $P(64.5 \leq X \leq 70.84+0.5=71.34)$ Calculate Z-scores: $$Z_1 = \frac{64.5 - 66.24}{4.3064} = \frac{-1.74}{4.3064} \approx -0.4043$$ $$Z_2 = \frac{71.34 - 66.24}{4.3064} = \frac{5.1}{4.3064} \approx 1.1846$$ From the standard normal table: $P(Z \leq -0.4043) \approx 0.3429$, $P(Z \leq 1.1846) \approx 0.8814$. So, $$P(64.5 \leq X \leq 71.34) = 0.8814 - 0.3429 = 0.5385$$ **Final answers (rounded to 4 decimal places):** (a) 0.5241 (b) 0.4759 (c) 0.5385