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The Medical College Admissions Test. Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). \({ }^{8}\) A new version of the exam was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 , the mean score was \(500.0\) with a standard deviation of \(10.6\). (a) What proportion of students taking the MCAT had a score over 510 ? (b) What proportion had scores between 505 and 515 ?

Short Answer

Expert verified
(a) 0.1721; (b) 0.2394

Step by step solution

01

Identify the Distribution

The problem indicates that MCAT scores are normally distributed, with a mean (\( \mu \)) of 500 and a standard deviation (\( \sigma \)) of 10.6.
02

Calculate Z-Score for Score 510

To find the proportion of students scoring above 510, we need the Z-score. The Z-score is calculated using the formula: \( Z = \frac{X - \mu}{\sigma} \). Here, \( X = 510 \). Therefore, \( Z = \frac{510 - 500}{10.6} \approx 0.943 \).
03

Find Proportion Above Z-Score for 510

Using the standard normal distribution table, find the probability that a Z-score is less than 0.943, which is approximately 0.8279. Therefore, the proportion of students scoring above 510 is \( 1 - 0.8279 = 0.1721 \).
04

Calculate Z-Scores for Scores 505 and 515

For 505: \( Z = \frac{505 - 500}{10.6} \approx 0.472 \). For 515: \( Z = \frac{515 - 500}{10.6} \approx 1.415 \).
05

Find Proportion Between Z-Scores for 505 and 515

Using the normal distribution table, find the probabilities: for 505 (\(Z = 0.472\)), it is approximately 0.6816, and for 515 (\(Z = 1.415\)), it is approximately 0.9210. The proportion between these Z-scores is \(0.9210 - 0.6816 = 0.2394 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
The normal distribution is a fundamental concept in statistics. It's a bell-shaped curve that is symmetrical around its mean. In this distribution, most values cluster around a central point, and the probabilities for values deviating from the mean taper off equally in both directions. This pattern is ideal for modeling many natural phenomena, including test scores, like those of the MCAT.
The key characteristics of a normal distribution include:
  • Symmetry: The left and right sides of the curve are mirror images.
  • Mean, median, and mode are all equal.
  • The curve is defined by two parameters: mean (\(\mu\)) and standard deviation (\(\sigma\)).
  • About 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three (the empirical rule).
Understanding the normal distribution helps predict probabilities for certain outcomes, as we did with the MCAT scores.
Z-Score Calculation
A Z-score is a measure that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations from the mean. Z-scores are essential for comparing different data points from diverse normal distributions.
To calculate a Z-score, you use the formula:\[Z = \frac{X - \mu}{\sigma}\]Where \(X\) is the value we are interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For instance, to find how a score of 510 compares to the MCAT mean score of 500, we calculate its Z-score, determining how many standard deviations 510 is from 500. This lets us use the normal distribution table to find probabilities.Calculating and understanding Z-scores is crucial for translating different scales and distributions into a standard form, making it possible to determine likelihoods and probabilities.
MCAT Scores Analysis
Analyzing MCAT scores involves understanding how scores are distributed among test-takers. The MCAT, or Medical College Admission Test, grades scores on a scale from 472 to 528, with a noted mean score of 500 and a standard deviation of 10.6. Such analysis helps examine how students perform relative to these statistics.
For practical analysis, we need to understand:
  • The mean score, which represents the average performance.
  • Standard deviation, indicating how much variation or dispersion from the average exists.
Knowing how to analyze these scores allows for insights into student performance and helps identify where a student stands in comparison to peers. It also facilitates determining the proportion of students scoring above a certain score—like 510—or within a specific range—such as between 505 and 515.
Standard Deviation
Standard deviation is a statistic that measures the dispersion of data points in a dataset relative to its mean. In simpler terms, it indicates how much individual scores vary from the average score. A small standard deviation means that most values are close to the mean, whereas a high standard deviation indicates that values are spread out over a wider range.
For the MCAT, with a standard deviation of 10.6, we can interpret:
  • A typical MCAT score will likely be within 10.6 points above or below the mean score of 500.
  • It helps to quantify the amount of variation or uncertainty in exam scores.
Overall, understanding standard deviation helps determine the reliability and predictability of the data distribution, useful for making informed assumptions about possible outcomes or when comparing different data sets.

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Most popular questions from this chapter

Grading managers. In Exercise 3.44, we saw that Ford Motor Company once graded its managers in such a way that the top \(10 \%\) received an A grade, the bottom \(10 \%\) a C, and the middle \(80 \%\) a B. Let's suppose that performance scores follow a Normal distribution. How many standard deviations above and below the mean do the \(\mathrm{A} / \mathrm{B}\) and \(\mathrm{B} / \mathrm{C}\) cutoffs lie? (Use the standard Normal distribution to answer this question.)

A surprising calculation. Changing the mean and standard deviation of a Normal distribution by a moderate amount can greatly change the percent of observations in the tails. Suppose a college is looking for applicants with SAT math scores 750 and above. (a) In 2015, the scores of men on the math SAT followed the \(N(527,124)\) distribution. What percent of men scored 750 or better? (b) Women's SAT math scores that year had the \(N(496,115)\) distribution. What percent of women scored 750 or better? You see that the percent of men above 750 is more than two and a half times the percent of women with such high scores. (On the other hand, women score higher than men on the new SAT writing test, though by a smaller amount.)

To completely specify the shape of a Normal distribution, you must give (a) the mean and the standard deviation. (b) the five-number summary. (c) the median and the quartiles.

Where are the quartiles? How many standard deviations above and below the mean do the quartiles of any Normal distribution lie? (Use the standard Normal distribution to answer this question.)

SAT versus ACT. In 2015 , when she was a high school senior, Linda scored 680 on the mathematics part of the SAT. \({ }^{5}\) The distribution of SAT math scores in 2015 was Normal with mean 511 and standard deviation 120 . Jack took the ACT and scored 26 on the mathematics portion. ACT math scores for 2015 were Normally distributed with mean \(20.8\) and standard deviation 5.4. Find the standardized scores for both students. Assuming that both tests measure the same kind of ability, who had the higher score? Men's and Women's Heights. The heights of women aged \(20-29\) in the United States are approximately Normal with mean \(64.2\) inches and standard deviation \(2.8\) inches. Men the same age have mean height \(69.4\) inches with standard deviation \(3.0\) inches. \({ }^{6}\) What are the \(z\)-scores for a woman \(5.5\) feet tall and a man \(5.5\) feet tall? Say in simple language what information the \(z\)-scores give that the original nonstandardized heights do not.

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