Statistical Problem Solving: Mean, Weighted Mean, and Score Adjustments, and Frequency Distributions

The primary objective of this problem type is to determine the mean score ( $\bar{x}$ ) when data is presented in a frequency distribution table rather than a raw list.
The Frequency Distribution Table Example: - Scores ( $x$ ): 1, 2, 3, 4, 5, 6 - Frequency ( $f$ ): 2, 0, 2, 4, 3, 1
Fundamental Formula for the Mean: - $\bar{x} = \frac{\sum x}{n}$ - The mean is the sum of all scores ( $\sum x$ ) divided by the sample size ( $n$ ).
Step-by-Step Procedure: - Step 1: Determine sample size ( $n$ ). Do NOT add up the $x$ column. To find the total number of people/scores, sum the frequency ( $f$ ) column. - $n = 2 + 0 + 2 + 4 + 3 + 1 = 12$ - Step 2: Determine the sum of all scores ( $\sum x$ ). You must create a new column representing the product of each score and its frequency ( $x \times f$ ). - $1 \times 2 = 2$ - $2 \times 0 = 0$ - $3 \times 2 = 6$ - $4 \times 4 = 16$ - $5 \times 3 = 15$ - $6 \times 1 = 6$ - Sum these products: $\sum x = 2 + 0 + 6 + 16 + 15 + 6 = 45$ - Step 3: Calculate the mean. - $\bar{x} = \frac{45}{12} = 3.75$
Reasonability Check: - The mean must fall within the range of the scores (between 1 and 6). - Given the distribution, a mean near the middle (3 or 4 point something) is expected. - Since $3.75$ is within this range and reflects the central tendency shown in the table, it is a reasonable and confident answer.

This problem involves combining two distinct samples to find the overall mean, a process known as the "weighted mean."
Setup Strategy: Create a table for Sample 1 and Sample 2, each with three variables: Mean ( $\bar{x}$ ), Sample size ( $n$ ), and Sum of scores ( $\sum x$ ). Filling in these six values provides all necessary data.
Weighted Mean Formula: - $\text{Weighted Mean} = \frac{\sum x_1 + \sum x_2}{n_1 + n_2}$
Example Problem Details: - Sample 1: Mean ( $\bar{x}_1$ ) = 8, Sample size ( $n_1$ ) = 10 - Sample 2: Mean ( $\bar{x}_2$ ) = 10, Sample size ( $n_2$ ) = 20
Algebraic Preparation: - To use the formula, we need the sum of scores for both samples. - For Sample 1: $8 = \frac{\sum x_1}{10} \rightarrow \sum x_1 = 80$ - For Sample 2: $10 = \frac{\sum x_2}{20} \rightarrow \sum x_2 = 200$
Calculation: - $\text{Weighted Mean} = \frac{80 + 200}{10 + 20} = \frac{280}{30} = 9.3333...$ - Following the class rounding rule of two decimal places: $9.33$
Predicting the Answer (The Weight Concept): - The mean must be between 8 and 10. - The sample with the larger size ( $n$ ) has more "weight" or "pull." - Since Sample 2 ( $n = 20$ ) is twice as large as Sample 1 ( $n = 10$ ), the weighted mean will be pulled closer to Sample 2's mean (10) and will be above the midpoint of 9.

This scenario occurs when an existing score in a sample is modified (e.g., a student retakes a quiz).
Example Data: - Original Sample ( $n$ ): 10 students - Original Mean ( $\bar{x}$ ): 16 - Change: One student's score went from 5 to 10.
Step 1: Find the original sum of scores ( $\sum x_{\text{before}}$ ). - $16 = \frac{\sum x_{\text{before}}}{10} \rightarrow \sum x_{\text{before}} = 160$
Step 2: Adjust the sum of scores ( $\sum x_{\text{after}}$ ). - Use the rigorous formula: $\sum x_{\text{after}} = \sum x_{\text{before}} - \text{old score} + \text{new score}$ - $\sum x_{\text{after}} = 160 - 5 + 10 = 165$
Step 3: Recalculate the mean. - The sample size ( $n$ ) remains 10 because a retake does not change the number of students. - $\text{New Mean} = \frac{165}{10} = 16.5$
Reasonability Comparison: Since a score went up by 5 points and the sample size is 10, it is logical that the class average increased slightly (from 16 to 16.5).

This problem determines what an individual must have scored to cause a specific shift in the mean when they are added to a sample.
Example Data: - Sample Before: $n = 10$ , Mean = 8.5 - Sample After: Mean = 8.0 - Change: One new student took the quiz late.
Step 1: Calculate the total points before the addition. - $\sum x_{\text{before}} = 8.5 \times 10 = 85$
Step 2: Calculate the total points after the addition. - The new sample size is $n = 11$ (10 students plus the 1 new student). - $\sum x_{\text{after}} = 8.0 \times 11 = 88$
Step 3: Determine the value of the added score. - $\text{Added Score} = \sum x_{\text{after}} - \sum x_{\text{before}}$ - $\text{Added Score} = 88 - 85 = 3$
Intuitive Indicator: The mean dropped from 8.5 to 8.0. This indicates the added score was lower than the original class average. The calculated score of 3 confirms this downward pull.

While there are sophisticated methods for calculating the median from a frequency distribution table, these will not be required for the current course testing.
Students should focus on finding the median from a simple list of scores as the primary required skill.