Scores on an exam for all medical school applications were approximately normal with a mean score of 503 and a standard deviation of 8.6. Suppose an applicant had a score of 512. What percentile corresponds with the score?
The Correct Answer and Explanation is:
To find the percentile that corresponds to a score of 512, given that exam scores are approximately normally distributed with a mean of 503 and a standard deviation of 8.6, we can follow these steps:
Step 1: Calculate the z-score
The z-score measures how many standard deviations a value is from the mean. The formula is:z=X−μσz = \frac{X – \mu}{\sigma}z=σX−μ
Where:
- X=512X = 512X=512 (the applicant’s score)
- μ=503\mu = 503μ=503 (the mean)
- σ=8.6\sigma = 8.6σ=8.6 (the standard deviation)
z=512−5038.6=98.6≈1.05z = \frac{512 – 503}{8.6} = \frac{9}{8.6} \approx 1.05z=8.6512−503=8.69≈1.05
Step 2: Find the percentile from the z-score
A z-score of 1.05 corresponds to a cumulative probability (or percentile) in the standard normal distribution table of approximately 0.8531.
This means that a score of 512 is higher than approximately 85.31% of the other scores.
Answer:
The percentile that corresponds with a score of 512 is approximately the 85th percentile.
Explanation
To interpret an exam score in the context of a normal distribution, we use the z-score. A z-score tells us how far a particular value lies from the mean of the distribution in terms of standard deviations. In this problem, the mean exam score is 503, and the standard deviation is 8.6. The applicant’s score of 512 is above the average, and calculating the z-score allows us to determine how unusual or competitive the score is.
After computing the z-score as approximately 1.05, we consult the standard normal distribution, which provides the area under the curve to the left of this z-score. This area represents the proportion of people who scored lower than 512. Looking up the value of 1.05 in the z-table gives approximately 0.8531, or 85.31%.
This tells us that the applicant scored better than about 85% of all test takers. The higher the z-score, the further to the right the value is in the distribution, indicating a better performance. Since the normal distribution is symmetric and bell-shaped, most scores lie near the mean, and scores far from the mean become increasingly rare. Thus, a z-score of 1.05 reflects a strong performance.
In real-world applications, percentile ranks like this help admission committees quickly evaluate how competitive a candidate’s score is relative to others. Being in the 85th percentile is considered quite good and shows that the applicant performed better than most of their peers.
