Leonel and Trent take entrance exams at two different universities. Leonel scores a 553 on an exam with a mean of 375 and a standard deviation of 64 . Leonel's corresponding z-score is Round your answer to two decimal places, if needed. Trent scores a 40 on an exam with á mean of 25 and a standard deviation of 4.5 . Round your answer to two decimal places, if needed. If acceptance is based on the entrance exam alone, who do you think is more likely to be accepted at the university of his choice? Leonel Trent

To calculate Leonel's z-score, use the formula: \( z = \frac{(X - \text{mean})}{\text{standard deviation}} \). Plugging in the values: \( z = \frac{(553 - 375)}{64} = \frac{178}{64} \approx 2.78 \). So, Leonel's z-score is approximately 2.78. For Trent, using the same formula: \( z = \frac{(40 - 25)}{4.5} = \frac{15}{4.5} \approx 3.33 \). Thus, Trent's z-score is approximately 3.33. Comparing their z-scores, Trent has a higher value (3.33) than Leonel (2.78), which suggests that Trent performed significantly better than his peers compared to Leonel. Therefore, Trent is more likely to be accepted at his university of choice. Both students did well, but Trent's higher z-score indicates he stands out more among his fellow test-takers!