Answer :
The percentage of students from the local high school who earn scores satisfying the admission requirement of the local college (minimum score of 2176) can be calculated by finding the area under the normal distribution curve beyond the z-score corresponding to the admission requirement. This percentage can be obtained by subtracting the cumulative probability from the mean of the distribution, converting it to a percentage.
To calculate the percentage of students meeting the admission requirement, we need to find the area under the normal distribution curve to the right of the z-score corresponding to the minimum score of 2176. This can be achieved by standardizing the minimum score using the z-score formula:
z = (x - μ) / σ
Where:
z is the z-score
x is the minimum score (2176)
μ is the mean of the distribution (1494)
σ is the standard deviation of the distribution (310)
Substituting the given values, we have:
z = (2176 - 1494) / 310
z ≈ 2.219
Next, we need to find the cumulative probability corresponding to this z-score. Using a standard normal distribution table or a calculator, we can find that the cumulative probability to the left of z = 2.219 is approximately 0.9857.
To find the percentage of students who earn scores satisfying the admission requirement, we subtract the cumulative probability from 1 (since we want the area to the right of the z-score) and convert it to a percentage:
Percentage = (1 - 0.9857) * 100
Percentage ≈ 1.4%
Therefore, approximately 1.4% of students from the local high school earn scores that satisfy the admission requirement of the local college.
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