Based on the given information, the cutoff score for the top 16% of applicants is 82.
How to find the cutoff score for the top 16% of applicants
To find the cutoff score for the top 16% of applicants, determine the z-score corresponding to this percentile and then convert it back to the original score using the mean and standard deviation.
Step 1: Find the z-score corresponding to the top 16% percentile.
Since the scores are normally distributed, use the standard normal distribution (z-distribution) to find the z-score.
The percentile corresponds to the area under the curve to the left of the z-score. We want to find the z-score that has an area of 0.16 to the left.
Using a standard normal distribution table or a calculator, the z-score corresponding to the top 16% percentile is approximately 0.99.
Step 2: Convert the z-score back to the original score using the mean and standard deviation.
z = (X - μ) / σ
Rearranging the equation, we have:
X = z * σ + μ
X = 0.99 * 13 + 69
X ≈ 81.87
The cutoff score, rounded up to the nearest whole number, is 82.
Therefore, the cutoff score for the top 16% of applicants is 82.