Problem 56 - An Introduction to Mathematical Statistics and Its Applications

Question

At State University, the average score of the entering class on the verbal portion of the SAT is 565 , with a standard deviation of 75 . Marian scored a 660 . How many of State's other 4250 freshmen did better? Assume that the scores are normally distributed.

Step-by-Step Solution

Answer

Approximately 434 students scored higher than Marian.

Step 1: Calculating the Z-score

First, calculate the z-score for Marian's SAT score. The z-score tells how many standard deviations an observation is from the mean. The formula is \( Z = (X - μ) / σ \), where X is the observation, μ is the mean, and σ is the standard deviation. Thus, \( Z = (660 - 565) / 75 \).

Step 2: Finding the percentile

Next, find the percentile for the calculated z-score. This can be done using a Standard Normal Distribution table. The percentile for a z-score of 1.27 on a standard normal distribution is approximately 0.8980. This means that approximately 89.8% of students scored lower than Marian did.

Step 3: Compute the number of students with higher score

The problem asks for the number of students who did better than Marian. So, we want the percentage of students who scored higher, not lower. Subtract the obtained percentile from 1 to get this percentage. So, \(1 - 0.8980 = 0.1020 \), or 10.20% of students have scores higher than 660. Multiply this percentage by the total number of students (4250) to get the count: \(0.1020 * 4250 ≈ 434\).

Key Concepts

Z-score

Understanding the Z-score is crucial in the realm of normal distribution. The Z-score represents the number of standard deviations that a data point, in this case, Marian's SAT score, is from the mean. To calculate it, we use the formula:

• \( Z = \frac{X - μ}{σ} \)

Where \( X \) is the observation (Marian's score of 660), \( μ \) is the mean score (565 for State University), and \( σ \) is the standard deviation (75). Applying these values, Marian's Z-score is calculated as:

• \( Z = \frac{660 - 565}{75} \approx 1.27 \)

A Z-score of 1.27 means that Marian scored 1.27 standard deviations above the average score. This enables comparisons across different data sets and can be used to determine Marian's position within the distribution of scores.

Standard Deviation

The Standard Deviation is a key measure in statistics, particularly in the context of a normal distribution. It quantifies the amount by which individual scores, such as SAT scores, differ from the mean. In this exercise, the standard deviation is 75, which tells us that typical variation in scores is 75 points from the average of 565. A smaller standard deviation indicates that scores are closer to the mean, whereas a larger one suggests wider variability.

The importance of knowing the standard deviation is that it helps in calculating the Z-score, which tells us how unusual or typical Marian's score is compared to her peers. Moreover, when examining a normal distribution, about 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three. Thus, in terms of student scores, most students at State University should have scores close to 565 ± 75, underscoring the concept of a 'normal' distribution.

Percentile Calculation

Percentile calculation is a common method used to understand how a particular score ranks in a normal distribution. Once Marian's Z-score (1.27) is calculated, we use it to find out what percentage of students scored below her. This involves using a standard normal distribution table, which correlates Z-scores with percentiles.

• A Z-score of 1.27 corresponds approximately to the 89.8th percentile.

This means that Marian did better than roughly 89.8% of her peers, implying that only about 10.2% scored higher than her.

To find the number of students who scored better, you multiply the percentage that scored higher by the total number of students (4250). Thus, about 10.2% of 4250 is 434 students. This conversion from Z-score to actual student count is not only part of the exercise but also a vital skill in numerous data analysis scenarios, where understanding rankings and relative performance are crucial.