Question

The test scores for a class have a normal distribution, a mean of \(50,\) and a standard deviation of 4. Quantity \(\mathbf{A}\) Percentage of scores at or above 58 Quantity \(\mathbf{B}\) Percentage of scores at or below 42 a. Quantity A is greater. b. Quantity \(\mathrm{B}\) is greater. c. The two quantities are equal. d. The relationship cannot be determined from the information given.

Short answer

The answer is c. The two quantities are equal.

Step-by-step solution

Determining score translate to z-score

As the first step, transform the scores 58 and 42 into z-scores. The formula to calculate z-score is: \(z = \frac{x - \mu}{\sigma}\) where x is the score, \(\mu\) is the mean and \(\sigma\) is the standard deviation. So, for 58 the z-score is \(z_{58} = \frac{58 - 50}{4} = 2\), and for 42 the z-score is \(z_{42} = \frac{42 - 50}{4} = -2\).

Using z-score to get percentage

Next, use the properties of normal distribution to find the percentages. For \(z_{58}=2\), the cumulative probability from mean to this z-score is about 0.9772, therefore the percentage of scores at or above 58 is \(1-0.9772 = 0.0228\) or 2.28%. Similarly, for \(z_{42}=-2\), it is already calculated as 0.0228, and the percentage of scores at or below 42 is the same value, 2.28%.

Comparing the quantities

Finally, compare Quantity A (Percentage of scores at and above 58) and Quantity B (Percentage of scores at and below 42). From Step 2, we know that both quantities amount to 2.28% of the scores. Thus, both quantities are equal.

Key concepts

Understanding Z-Score Calculation

When dealing with a normal distribution, the concept of z-score is essential. A z-score is a standardized way of expressing the position of a single data point relative to the mean of the data set. It tells us how many standard deviations away a particular score is from the average.

The formula for calculating a z-score is:

• \( z = \frac{x - \mu}{\sigma} \)

Here, \(x\) is the data point in question, \(\mu\) is the mean of the data set, and \(\sigma\) is the standard deviation.

In our exercise, we translated a test score of 58 into a z-score by substituting into the formula, giving us \(z_{58} = \frac{58 - 50}{4} = 2\). Similarly, for a score of 42, we used \(z_{42} = \frac{42 - 50}{4} = -2\). This calculation tells us that the score of 58 is 2 standard deviations above the mean and the score of 42 is 2 standard deviations below the mean.

Delving Into Mean and Standard Deviation

Mean and standard deviation are foundational statistics concepts that provide a grasp on the "center" and "spread" of a data set.

- **Mean (\(\mu\))**: This is simply the average of all data points in a set. In our context, the mean score of the test is 50.

- **Standard Deviation (\(\sigma\))**: This tells us how much the scores deviate from the mean on average. A small standard deviation implies the scores are tightly clustered around the mean, whereas a large one suggests a wider spread. Here, the standard deviation is 4.

Knowing these values, we can determine how unusual a particular score is by using the z-score. For example, a score of 58 has a positive z-score (above the mean), whereas a score of 42 has a negative z-score (below the mean). This understanding assists in interpreting the distribution reliably.

Exploring the Concept of Percentage Comparison

After obtaining z-scores, the next step involves translating these into meaningful probabilities, or percentages, using the properties of the normal distribution.

For the exercise, we aimed to determine:

• The percentage of scores at or above 58.
• The percentage of scores at or below 42.

Both scores correspond to z-scores of \(z = 2\) and \(z = -2\), respectively.

To find these percentages, we look at the cumulative probability up to those z-scores. For \(z = 2\), the cumulative probability from the mean to this score is approximately 0.9772. Therefore, the percentage at or above 58 is \(1 - 0.9772 = 0.0228\), or 2.28%.
Conversely, for \(z = -2\), the situation is mirrored, giving 2.28% for scores at or below 42. With both percentages computed, comparing these quantitatively shows they are equal at 2.28% each, assisting in solving the exercise entirely.