1. **State the problem:** Andy believes 83% of his customers are satisfied with the food. A random sample of 99 customers is taken. We want to find probabilities related to the number and proportion of satisfied customers. 2. **Identify parameters:** - Population proportion $p=0.83$ - Sample size $n=99$ - Number of successes $X$ follows a binomial distribution: $X \sim \text{Binomial}(n=99, p=0.83)$ 3. **Use normal approximation to binomial** (since $np=99 \times 0.83=82.17$ and $n(1-p)=99 \times 0.17=16.83$ are both > 5): - Mean $\mu=np=82.17$ - Standard deviation $\sigma=\sqrt{np(1-p)}=\sqrt{99 \times 0.83 \times 0.17} \approx \sqrt{13.963} = 3.7379$ 4. **Part a) Probability less than 83 customers satisfied:** Use continuity correction: $P(X < 83) = P(X \leq 82)$, approximate as $P(Y \leq 82.5)$ where $Y$ is normal. Calculate z-score: $$z=\frac{82.5 - 82.17}{3.7379} = \frac{0.33}{3.7379} \approx 0.09$$ Find probability from z-table: $$P(Z \leq 0.09) \approx 0.5359$$ 5. **Part b) Probability at least 83 customers satisfied:** $$P(X \geq 83) = 1 - P(X \leq 82) = 1 - P(Y \leq 82.5)$$ From part a), $$= 1 - 0.5359 = 0.4641$$ 6. **Part c) Probability sample proportion between 80% and 88%:** Translate proportions to counts: - Lower bound: $0.80 \times 99 = 79.2$ - Upper bound: $0.88 \times 99 = 87.12$ Using continuity correction: $$P(79.2 < X < 87.12) \approx P(79.5 \leq Y \leq 87.5)$$ Calculate z-scores: $$z_1= \frac{79.5 - 82.17}{3.7379} = \frac{-2.67}{3.7379} \approx -0.71$$ $$z_2= \frac{87.5 - 82.17}{3.7379} = \frac{5.33}{3.7379} \approx 1.43$$ Find probabilities from z-table: $$P(Z \leq 1.43) \approx 0.9236$$ $$P(Z \leq -0.71) \approx 0.2389$$ Calculate final probability: $$0.9236 - 0.2389 = 0.6847$$ **Final answers:** a) 0.5359 b) 0.4641 c) 0.6847