The combined math and verbal scores for students taking a national standardized examination for college admission is normally distributed with a mean of 560 and a standard deviation of 180. If a college requires a student to be in the top 40 % of students taking this test, what is the minimum score that such a student can obtain and still qualify for admission at the college?
To find the minimum score that a student needs to qualify for admission at the college, we need to determine the score corresponding to the top 40% of students.
First, let's find the z-score corresponding to the top 40% of the distribution using the standard normal distribution table (also known as the z-score table) or a statistical calculator. The top 40% represents the area to the right of the score we are looking for.
Using the formula for z-score:
z = (x - μ) / σ
where:
x = the score we are looking for
μ = the mean of the distribution (560)
σ = the standard deviation of the distribution (180)
To convert the top 40% to a z-score, we look for the z-score corresponding to 1 - 0.40 = 0.60 in the standard normal distribution table.
The z-score corresponding to 0.60 is approximately 0.25 (you can verify this using a z-score table or calculator).
Now, we can rearrange the formula for z-score to solve for the score x:
x = z * σ + μ
x = 0.25 * 180 + 560
x = 45 + 560
x = 605
Therefore, the minimum score a student needs to obtain to qualify for admission at the college is 605.