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Sagot :
We need o find the cut-off score for acceptance.
We know that the scores are normally distributed with a mean of 500 and a standard deviation of 100.
Thus, we can use a z-score table to find Z for which the percentage above it is 25% = 0.25.
Then, we calculate the cut-off score x as follows:
[tex]z=\frac{x-\text{ mean}}{\text{ standard deviation}}[/tex]Using a z-score table, we find the the z with a percentage above 0.25 (one minus the percentage below 0.75) is:
[tex]z\cong0.674[/tex]Then, we obtain:
[tex]\begin{gathered} 0.674=\frac{x-500}{100} \\ \\ 67.4=x-500 \\ \\ x=67.4+500 \\ \\ x=567.4 \\ \\ x\cong567 \end{gathered}[/tex]Answer: c) 567
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