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 B is greater. c. The two quantities are equal. d. The relationship cannot be determined from the information given.

Short answer

c. The two quantities are equal.

Step-by-step solution

Calculation of Z-scores

Start by calculating the Z-scores for the score 58 and 42. The Z-score formula is given by \(Z = (X - µ)/σ\), where X represents a given score, µ is the mean score, and σ is the standard deviation. For 58, it is \(Z_1 = (58-50)/4 = 2\) and for 42, it is \(Z_2 = (42-50)/4 = -2\)

Reference to Z-table

With the obtained Z-scores, the next step is referring to the standard normal distribution table (Z-table). The Z-table gives the proportion of scores that fall to the left of a specified Z-score in a standard normal distribution. Given that Z_1=2, the value corresponding to Z=2 in the Z-table is 0.9772. This means the proportion of scores less than 58 is 0.9772 or 97.72%. Now, for Z_2=-2, the Z-table gives the value as 0.0228 which represents the proportion of scores less than 42. However, we want the proportion below 42, and the Z-score of -2 maps to 0.0228 in the Z-table, it means that the proportion of scores less than or equal to 42 is 0.0228 or 2.28%.

Derivation of the Quantities

For Quantity A, we need to find the percentage of scores at or above 58. Since we found the percentage below 58 as 97.72% using the Z-table, the percentage at or above 58 will be \(100% - 97.72% = 2.28%\). For Quantity B, we need the percentage of scores at or below 42. As we found from the Z-table, this percentage is 2.28%.

Comparison of the Quantities

Now we can compare Quantity A and Quantity B. Both Quantities A and B equate to 2.28%. Hence, the two quantities are equal.

Key concepts

Z-scores

When dealing with normal distributions, Z-scores play a crucial role. A Z-score tells us how many standard deviations a particular score is from the mean. This score allows us to determine where a value falls within a normal distribution.

To calculate a Z-score, you use the formula:

• \( Z = \frac{X - \mu}{\sigma} \)

Here, \( X \) represents the individual score being analyzed, \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation.
For example, in our problem, we calculated the Z-scores for scores 58 and 42 based on a mean (\(\mu\)) of 50 and a standard deviation (\(\sigma\)) of 4.

• For score 58: \( Z_1 = \frac{58 - 50}{4} = 2 \)
• For score 42: \( Z_2 = \frac{42 - 50}{4} = -2 \)

These Z-scores help us find the position of these scores in the normal distribution and are essential for using the normal distribution table.

Standard Deviation

Standard deviation is a measure of how spread out numbers are in a data set. It gives us an idea of the variation or dispersion in the distribution of scores. A small standard deviation indicates that the scores are close to the mean, while a larger standard deviation suggests that the scores are more spread out.

In the context of a normal distribution, the standard deviation determines the shape and spread of the bell curve. For example, if a test score distribution has a mean of 50 and a standard deviation of 4, as in our problem, most of the scores will be within a few standard deviations of the mean.

• A score within 1 standard deviation from the mean (between 46 and 54) includes about 68% of the data.
• Within 2 standard deviations from the mean (between 42 and 58), about 95% of the data is included.
• Within 3 standard deviations covers about 99.7% of the data.

Understanding standard deviation helps in predicting probabilities and making inferences from data in a normal distribution.

Normal Distribution Table

The normal distribution table, often called the Z-table, is a valuable tool when dealing with normally distributed data. It allows you to find the proportion of data points below, above, or between given Z-scores.

To use a Z-table:

• Identify the Z-score you are interested in.
• Locate the corresponding value in the table, which represents the percentage of data below that Z-score.

In our example, after calculating Z-scores of 2 for a score of 58 and -2 for a score of 42:

• The Z-table shows that 97.72% of scores are below a Z-score of 2.
• For a Z-score of -2, only 2.28% of scores are below this value.

These percentages help us understand the position and range of the data within the normal distribution. The Z-table provides a standardized way to interpret data that follows a normal distribution.

Mean Score Calculation

Calculating the mean score is the foundational step in understanding a data set's center. The mean score is essentially the arithmetic average of the scores and is represented by the symbol \(\mu\).

To calculate the mean:

• Add up all the scores.
• Divide the total by the number of scores.

In a normally distributed data set, the mean is the peak or center of the bell curve. It is the point where half of the observations fall below and half above, ensuring that the distribution is symmetric.
For our specific problem, the mean was provided as 50. This value is critical as it anchors the Z-score calculations by providing a benchmark for interpreting how much a particular score deviates from this central tendency.
Understanding the mean assists in various statistical analyses and provides a reference point for calculating measures like the standard deviation and determining Z-scores.