Answer :
To find the cutoff score for the top 10% of test scores for college admission, we use the z-score corresponding to the 90th percentile in a normal distribution, which is approximately 1.28. Using the formula X = Zσ + μ with a mean of 625 and standard deviation of 22, we find the cutoff score to be approximately 653.
If an elite private college wants to admit students whose test scores are in the top 10% of the scores, we need to find the cutoff score that corresponds to this percentile in a normal distribution with a mean score of 625 and a standard deviation of 22.
To do this, we would use a standard normal distribution table or z-score calculator to find the z-score that corresponds to the 90th percentile since the normal distribution is symmetric, and the top 10% would begin at the 90th percentile. To find the z-score for the 90th percentile, we look up the value in a standard normal distribution table or use a calculator, which gives us a z-score of approximately 1.28. Then, we use the z-score formula:
- Z = (X - \\(\mu\\)) / \\(\sigma\\),
- where X is the cutoff score, \\\(\mu\\) is the mean, and \\\(\sigma\\) is the standard deviation. Solving for X gives us:
- X = Z \\\(\sigma\\) + \\\(\mu\\) = 1.28(22) + 625 = 28.16 + 625 = 653.16.
Therefore, the cutoff score to be in the top 10% should be approximately 653.
