Question

A group of students at a school takes a history test. The distribution is normal with a mean of \(25,\) and a standard deviation of \(4 .\) (a) Everyone who scores in the top \(30 \%\) of the distribution gets a certificate. What is the lowest score someone can get and still earn a certificate? (b) The top \(5 \%\) of the scores get to compete in a statewide history contest. What is the lowest score someone can get and still go onto compete with the rest of the state?

Short answer

Students need at least 27.1 for a certificate and 31.6 to compete statewide.

Step-by-step solution

Understand the Problem

We need to find the score thresholds for the top 30% and top 5% in a normally distributed test with a mean of 25 and standard deviation of 4.

Use Z-Score Formula

For a normal distribution, the formula to find the Z-score corresponding to a probability is: \[ Z = \frac{X - \mu}{\sigma} \] where \(X\) is the score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Find Z-Score for Top 30%

For the top 30%, we need the Z-score where 70% of the data lies below it. Using a Z-table or calculator, the Z-score for 0.70 is approximately 0.524.

Calculate the Lowest Score for Top 30%

Using the Z-score formula, substitute the values: \[ X = Z \cdot \sigma + \mu = 0.524 \cdot 4 + 25 = 27.096 \] So, the lowest score to earn a certificate is approximately 27.1.

Find Z-Score for Top 5%

For the top 5%, we need the Z-score where 95% of the data lies below it. Using a Z-table or calculator, the Z-score for 0.95 is approximately 1.645.

Calculate the Lowest Score for Top 5%

Using the Z-score formula, substitute the values: \[ X = Z \cdot \sigma + \mu = 1.645 \cdot 4 + 25 = 31.58 \] So, the lowest score to compete in the statewide contest is approximately 31.6.

Key concepts

Understanding Z-Scores

In the world of statistics, the z-score is a crucial concept that helps us understand where a particular value lies within a normal distribution. The z-score represents the number of standard deviations a data point is from the mean. It's essentially a way to transform any normal-shaped data into a standard normal distribution, which has a mean of 0 and a standard deviation of 1.

For example, if a z-score is positive, it indicates that the data point is above the mean. Meanwhile, a negative z-score signifies a point below the mean. A z-score of zero implies the value is exactly at the mean.

• The formula to calculate the z-score is \( Z = \frac{X - \mu}{\sigma} \), where:
  
• \( X \) is the data point.
  
• \( \mu \) is the mean of the distribution.
  
• \( \sigma \) is the standard deviation.

By converting test scores into z-scores, you can quickly gauge how students' performances compare to the average, helping determine thresholds for awards or contests.

Mean and Standard Deviation Basics

The mean and standard deviation are fundamental elements in understanding a normal distribution. In simple words, the mean is the average of all the numbers in a dataset. It gives you a general idea about the data's central tendency or where most values cluster.

On the other hand, the standard deviation is a little different. It tells you how spread out the numbers are from the mean. A smaller standard deviation means the values tend to be closer to the mean, while a larger one suggests a wider spread of numbers.

• To find the mean (\( \mu \)), sum all the data points and divide by the total number of points.
  
• The standard deviation (\( \sigma \)) requires a few more steps:
  • First, calculate the variance by finding the average of the squared differences from the mean.
    
  • Then take the square root of the variance to get the standard deviation.
    

When solving problems involving normal distributions, knowing the mean and standard deviation allows you to calculate z-scores with ease, helping you analyze the data efficiently.

Navigating Probability Thresholds

Probability thresholds are a way to set benchmarks within a distribution to decide critical cut-offs, such as determining the top-performing individuals in a test. In our exercise, the goal is to pinpoint the scores that separate the top 30% and top 5% of test-takers.

• To make these decisions, the z-score approach is invaluable. By using a z-table or calculator, you find the z-score that corresponds to your desired probability percentile length.
  
• For instance, a 70% percentile means finding the z-score where 70% of data lies below. This z-score can then be translated back into an actual score with the formula \( X = Z \cdot \sigma + \mu \).
  
• This process helps establish concrete score thresholds, like 27.1 for the top 30% and 31.6 for the top 5% participants.

Establishing these thresholds helps recognize outstanding achievers and is essential in academic settings for awarding recognitions or selecting candidates to advance to competitions.