Given,
The mean of the test score is 1013.
The standard deviation of the test score is 125.
The percentage of cutoff is 9%.
The z score at 9% is,
[tex]Z=1.35[/tex]
The z score is calculated as,
[tex]z=\frac{X-\operatorname{mean}}{s\tan dard\text{ deviation}}[/tex]
Substituting the value of z, mean and standard deviation then,
[tex]\begin{gathered} 1.35=\frac{X-1013}{125} \\ 168.75=X-1013 \\ X=1182 \end{gathered}[/tex]
Hence, the cutoff for the top 9% is 1182.
