The variable of interest is
X: number of students that use a pencil to take notes out of 120.
This variable has a binomial distribution with parameters n=120 and p. To calculate the confidence interval for the population proportion of students that use a pencil to take notes you have to use the approximation of the standard normal distribution with margin of error:
[tex]Z_{\mleft\lbrace1-\frac{\alpha}{2}\mright\rbrace}\sqrt[]{\frac{p^{\prime}(1-p^{\prime})^{}}{n}}[/tex]The sample proportion is the number of successes of the experiment divided the sample size, in this case the successes are the students that use a pencil so the sample proportion is:
p'=90/120=0.75
From the margin of error the only value modified will ve the Z value, so calculate the term under the square root first:
1-p'=1-0.75=0.25
[tex]\sqrt[]{\frac{0.75\cdot0.25}{120}}=\frac{\sqrt[]{10}}{80}\cong0.0395[/tex]Next, using the Z-table look for the corresponding values:
1) For 1-α= 99%=0.99
α=0.01
α/2=0.005
1-α/2=0.995
[tex]Z_{\mleft\lbrace1-\frac{\alpha}{2}\mright\rbrace}=Z_{\mleft\lbrace0.995\mright\rbrace}=2.576[/tex]2) For 1-α=95%=0.95
α=0.05
α/2=0.025
1-α/2=0.975
[tex]Z_{\mleft\lbrace0.975\mright\rbrace}=1.960[/tex]3) For 1-α=90%=0.90
α=0.1
α/2=0.05
1-α/2=0.95
[tex]Z_{\mleft\lbrace0.95\mright\rbrace}=1.645[/tex]