Answer :
To determine the percentage of students who meet the admission requirement for the local college, we start by understanding that the SAT scores are normally distributed with a mean ([tex]\mu[/tex]) of 1493 and a standard deviation ([tex]\sigma[/tex]) of 303.
The local college requires a minimum SAT score of 675 for admission. We need to find the percentage of students who score at least 675.
Step-by-Step Solution:
Calculate the z-score:
The z-score is calculated using the formula:
[tex]z = \frac{X - \mu}{\sigma}[/tex]
Where:
- [tex]X[/tex] is the score we are comparing, 675.
- [tex]\mu[/tex] is the mean, 1493.
- [tex]\sigma[/tex] is the standard deviation, 303.
[tex]z = \frac{675 - 1493}{303}[/tex]
[tex]z \approx \frac{-818}{303}[/tex]
[tex]z \approx -2.70[/tex]Use the z-score to find the percentage:
A z-score of [tex]-2.70[/tex] means that 675 is [tex]2.70[/tex] standard deviations below the mean. Using a standard normal distribution table or a calculator, we look up the probability that corresponds to [tex]z = -2.70[/tex].
This probability tells us the percentage of students scoring below 675. For [tex]z = -2.70[/tex], the probability is approximately 0.35%.
Calculate the students who meet the requirement:
Since only 0.35% score below 675, the remaining percentage of students score 675 or above.
[tex]100\% - 0.35\% \approx 99.65\%[/tex]
Conclusion:
Approximately 99.65% of students from this high school earn scores that satisfy the admission requirement of scoring at least 675 on the SAT. Therefore, almost all students meet the college's minimum admission score requirement.
