Answer :
Final answer:
To be in the top 15% of a normally distributed admission test, a candidate would typically need a score of about one standard deviation above the mean score. The exact 'cutoff' value depends on the specifics of the distributions, such as the mean and standard deviation.
Explanation:
The question is inquiring about the minimum score needed to be among the top 15 percent on a test, assuming that the test scores are normally distributed. This requires a bit of understanding of the properties of a normal distribution. In a normal distribution, 68 percent of the scores are within one standard deviation from the mean, 95 percent are within two standard deviations, and nearly all (99.7 percent) are within three standard deviations. This is often referred to as the 68-95-99.7 rule.
A common significance level used in hypothesis testing is 5 percent, which often generally corresponds to being in the top 5 percent or the bottom 5 percent of distribution. However, in this case, we are interested in being in the top 15 percent.
If scores were distributed normally, 50 percent would be above the mean and 50 percent below. The top 15 percent in a normal distribution usually falls around 1 standard deviation above the mean. So, you'd look at the mean value of admitted students' test scores, and the standard deviation of those scores, and add that standard deviation to the mean to find the 'cutoff.' However, this depends on the exact characteristics of the scores' distribution.
So without specific numbers for the mean and standard deviation, we cannot give a concrete number for the minimum score needed to be in the top 15 percent. However, the process would be as described.
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