1. Problem statement. Andy believes that 76% of his customers are satisfied with the food quality, and a random sample of 84 customers is taken to estimate probabilities for counts and proportions. We are asked to find: (a) the probability that fewer than 64 customers are satisfied, (b) the probability that at least 64 are satisfied, and (c) the probability that the sample proportion satisfied is between 0.73 and 0.80. 2. Model and parameters. Model the number satisfied $X$ as $\text{Binomial}(n,p)$ with $n=84$ and $p=0.76$. The mean of $X$ is $$\mu=np=84\cdot0.76=63.84$$ The standard deviation of $X$ is $$\sigma=\sqrt{np(1-p)}=\sqrt{84\cdot0.76\cdot0.24}=\sqrt{15.3216}\approx3.9143$$ We may use the normal approximation with continuity correction because $n$ is large and $np(1-p)$ is well above 5. 3. Part (a): $P(X<64)$. The event $X<64$ is equivalent to $X\le63$, so use continuity correction and compute $$Z=\frac{63.5-\mu}{\sigma}=\frac{63.5-63.84}{3.9143}=\frac{-0.34}{3.9143}\approx-0.0869$$ Using the standard normal CDF $\Phi$, the probability is $$P(X<64)\approx\Phi(-0.0869)\approx0.4654$$ Therefore the probability that fewer than 64 customers are satisfied is 0.4654. 4. Part (b): $P(X\ge64)$. This is the complement of $X\le63$, so $$P(X\ge64)=1-P(X\le63)\approx1-0.4654=0.5346$$ Therefore the probability that at least 64 customers are satisfied is 0.5346. 5. Part (c): $P(0.73\le\hat p\le0.80)$ where $\hat p=X/n$. The sampling distribution of $\hat p$ has mean $p=0.76$ and standard error $$\text{SE}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.76\cdot0.24}{84}}=\sqrt{0.0021714286}\approx0.04659$$ Compute the z-scores for the endpoints: $$z_{\text{low}}=\frac{0.73-0.76}{0.04659}\approx-0.6440$$ $$z_{\text{high}}=\frac{0.80-0.76}{0.04659}\approx0.8587$$ Using the standard normal CDF values, $$P(0.73\le\hat p\le0.80)\approx\Phi(0.8587)-\Phi(-0.6440)\approx0.8047-0.2598\approx0.5449$$ Therefore the probability that the sample proportion is between 73% and 80% is 0.5449. Final answers (rounded to four decimal places): (a) 0.4654 (b) 0.5346 (c) 0.5449