Answer
p is between 0.92-0.1027=0.8173 and 0.92+0.1027=1.0227. The thing that is unusual that the upper bound exceeds 1, which is not possible for p. Modification is needed.
Work Step by Step
The best point estimate is equal to the proportion of the sample (x) divided by the sample size: $\hat{p}=\frac{x}{n}=\frac{44}{48}=0.92.$
b)$E=z_{\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat{p}\cdot (1-\hat{p})}{n}}=2.575\cdot \sqrt{\frac{0.92\cdot (1-0.92)}{48}}=0.1027.$
Hence, the confidence interval: E is between $\hat{p}-E$ and $\hat{p}+E$, hence p is between 0.92-0.1027=0.8173 and 0.92+0.1027=1.0227. The thing that is unusual that the upper bound exceeds 1, which is not possible for p. Hence, modification is needed.
