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Sagot :
Given
Mean is 500 and Standard Deviation is 100.
Part A
Using Z- Score
The minimum score necessary to be in the top 10% of the SAT distribution means 90% and above
[tex]\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ \\ x=0.9 \\ \mu=500 \\ \sigma=100 \\ 1.282=\frac{x-500}{100} \\ \\ x=628.2 \end{gathered}[/tex]Thus, the minimum score is 628. To the nearest whole number.
Part B
We want to determine the range of values that correspond to the probability;
[tex]Pr(-ZThe z scores corresponding to this is; [tex]z=+1.282\text{ }&-1.282[/tex]Thus;
[tex]\begin{gathered} 1.282=\frac{x-500}{100}\text{ and }-1.282=\frac{x-500}{100} \\ 628.2=x\text{ }and\text{ }371.8 \end{gathered}[/tex]Thus, the range of scores are 372 and 628
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