the average act score submitted to a certain university is 26. to be eligible for admission, a student must score in the top 6% of applicants. if the scores are normally distributed with a standard deviation of 2.4, what is the minimum score needed to be considered for admission?
Z = (x -26)/2.4
1.55 = (x-26)/2.4
1.55(2.4) = x -26
1.55(2.4) + 26 = x
To find the minimum score needed to be considered for admission, we need to determine the cutoff score that separates the top 6% of applicants from the rest.
Given that the average ACT score submitted is 26 and the scores are normally distributed with a standard deviation of 2.4, we can use z-scores to find the cutoff score.
A z-score measures how many standard deviations a particular value is from the mean. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
- x is the value in question (the cutoff score)
- μ is the mean (average ACT score)
- σ is the standard deviation
We need to find the z-score corresponding to the top 6%. Since the distribution is symmetric, the top 6% is the same as the bottom 6%.
Using a standard normal distribution table or a calculator, we can find that the z-score corresponding to the bottom 6% is approximately -1.556.
Now, we can rearrange the z-score formula to solve for x:
-1.556 = (x - 26) / 2.4
Next, let's simplify the equation:
-1.556 * 2.4 = x - 26
-3.74 = x - 26
x = 26 - 3.74
x ≈ 22.26
Therefore, the minimum ACT score needed to be considered for admission is approximately 22.26.