Question

last year over 13,000 students took an entrance exam at a certain state university. janes score was at the 57th percentile. joses score was at the 39th percentile.
(a) which of the following must be true about janes and joses scores?
both jane and jose scored higher than the median.
both jane and jose scored below than the median.
jane scored higher than jose.
jose scored higher than jane.
(b) which of the following must be true about janes score?
about 57% of the students who took the exam scored lower than jane.
jane got about 57% of the questions correct.
janes score was in the bottom half of all scores.
jane missed 43 questions.

Explanation:

(a)

• Understanding Percentiles:
• The median is the \(50^{th}\) percentile. Jane's score is at the \(57^{th}\) percentile (since \(57>50\)), so Jane scored above the median. Jose's score is at the \(39^{th}\) percentile (since \(39 < 50\)), so Jose scored below the median. So the first two options (both above or both below the median) are wrong.
• Since \(57>39\) (because if \(x\) is at the \(p^{th}\) percentile and \(y\) is at the \(q^{th}\) percentile, and \(p>q\), then \(x\) scored higher than \(y\) when the data is ordered from low - to - high).

(b)

• Interpretation of Percentile Definition:
• By the definition of percentile, if a score is at the \(p^{th}\) percentile, approximately \(p\%\) of the data values are less than that score. So if Jane's score is at the \(57^{th}\) percentile, about \(57\%\) of the students who took the exam scored lower than Jane.
• The percentile does not tell us the proportion of questions correct (it's about the relative position among all test - takers, not the accuracy of answering questions), so the option “Jane got about \(57\%\) of the questions correct” is wrong. Since \(57>50\), Jane's score was not in the bottom half. And we have no information about the number of questions (to say she missed 43 questions).

Answer:

(a) Jane scored higher than Jose.
(b) About \(57\%\) of the students who took the exam scored lower than Jane.