The variable of interest is X: the number of people that answered positively when asked if they felt vulnerable to identity theft, out of 1100.
This variable has a binomial distribution.
To calculate a confidence interval for the population proportion of people that answered "yes" to the poll, you have to use the approximation to the standard normal distribution:
![Z=\frac{p\lbrack hat\rbrack-p}{\sqrt[]{(p\lbrack hat\rbrack(1-p\lbrack hat\rbrack))/(n)}}\approx N(0,1)](https://img.qammunity.org/2023/formulas/mathematics/college/ey8h3sqmn5vodiyy00swvent6bk30x0ymo.png)
The structure for the formula of the confidence interval is "estimator"±"margin of error"
Where the estimator is the sample proportion p[hat] and the margin of error has the following form:
![Z_{\mleft\lbrace1-(\alpha)/(2)\mright\rbrace}\sqrt[]{(p\lbrack hat\rbrack(1-p\lbrack hat\rbrack))/(n)}](https://img.qammunity.org/2023/formulas/mathematics/college/6398x9nexlluiwiel90yjbot287bu05d3r.png)
To calculate the margin of error you have to determine the Z-value and the value of the sample proportion
Z-value, determine the probability, and then look for the value on the Z-table:
Confidence level: 1-α= 0.90
α=0.1
α/2=0.05

The sample proportion can be calculated by dividing the number of successes, in this case, the number of people that answered "yes" by the total number of people surveyed:

With these values you can determine the margin of error of the confidence interval as follows:
![\begin{gathered} Z_{\mleft\lbrace1-(\alpha)/(2)\mright\rbrace}\sqrt[]{(p\lbrack hat\rbrack(1-p\lbrack hat\rbrack))/(n)} \\ 1.645\cdot\sqrt[]{((542)/(1100)(1-(542)/(1100)))/(1100)=0.0249\approx0.025} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h95qiuh5jn2k4ac939nm13f9gg8hvycrqw.png)
The margin of error of the 90% confidence interval for the population proportion is 0.025
Confidence interval:
