Answer
p is between 0.92-0.0656=0.8544 and 0.92+0.0656=0.9856. No.
Work Step by Step
The best point estimate is equal to the proportion of the sample (x) divided by the sample size: $\hat{p}=\frac{x}{n}=\frac{44}{48}=0.92.$
$E=z_{\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat{p}\cdot (1-\hat{p})}{n}}=1.645\cdot \sqrt{\frac{0.92\cdot (1-0.92)}{48}}=0.0656.$
Hence, the confidence interval: E is between $\hat{p}-E$ and $\hat{p}+E$, hence p is between 0.92-0.0656=0.8544 and 0.92+0.0656=0.9856.
The confidence interval doesn't describe the percentage of all on-time American Airlines flights, because the percentage needn't be in the interval.
