Answer :
Final answer:
The Medical College Admission Test (MCAT) is required for admission to many U.S. medical schools. Scores on the MCAT are normally distributed with a mean of 25.0 and a standard deviation of 6.4. To calculate the z-score for a specific MCAT score, you can use the formula: z = (x - µ) / σ
Explanation:
The Medical College Admission Test (MCAT) is required for admission to many U.S. medical schools. The scores on the MCAT are normally distributed with a mean of 25.0 and a standard deviation of 6.4. This means that most students will receive scores close to the mean, with fewer students receiving higher or lower scores.
To calculate the z-score for a specific MCAT score, you can use the formula:
z = (x - ) /
where x is the individual MCAT score, is the mean, and is the standard deviation.
The mean score of 25.0 represents the average performance on the MCAT, while the standard deviation of 6.4 indicates the spread of scores around the mean.
The Medical College Admission Test (MCAT) is a required exam for admission to many medical schools in the United States. MCAT scores follow a normal distribution with a mean of 25.0 and a standard deviation of 6.4.
In a normal distribution, the majority of scores cluster around the mean, with fewer scores farther away. This distribution allows medical schools to evaluate applicants' performance relative to other test takers. The mean score of 25.0 represents the average performance on the MCAT, while the standard deviation of 6.4 indicates the spread of scores around the mean.
The MCAT is a standardized exam that assesses an individual's knowledge of scientific concepts, critical thinking skills, and problem-solving abilities necessary for success in medical school. The normal distribution of MCAT scores means that most test takers fall near the mean score of 25.0.
The standard deviation of 6.4 indicates the average amount of variability or dispersion of scores from the mean. This implies that approximately 68% of test takers will have scores within one standard deviation of the mean (between 18.6 and 31.4), while around 95% will have scores within two standard deviations (between 12.2 and 37.8).
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