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Sagot :

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.

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