Pregunta
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Calculate the mean, the median, and the mode for each of the following data sets. a. \( 3,7,7,8,6,7,12,4 \) b. \( 10.5,12.5,12.5,14.5,19.5,16.5,12.5,14.5,11.5 \) c. \( 5,-4.2,-5.4,6.2,0,3.2,10.3,0.9 \) d. \( 74,79,87,79,73,76,5 \) Select the correct choice below and fill in any answer Joxes within your choice. A. The mode(s) is/are 7 . (Use a comma to separate answers as needed.) B. There is no mode. b. The mean is \( \square \). (Round to the nearest thousandth as needed.)

Ask by Gibbs Bradley. in the United States
Feb 03,2025

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- **A. The mode(s) is/are 7.** - **b. The mean is \( 6.750 \).**

Solución

Let's analyze **Problem a**: \( 3, 7, 7, 8, 6, 7, 12, 4 \). ### Step 1: Calculate the Mean The mean is the average of all the numbers in the dataset. \[ \text{Mean} = \frac{\sum \text{values}}{\text{number of values}} = \frac{3 + 7 + 7 + 8 + 6 + 7 + 12 + 4}{8} = \frac{54}{8} = 6.75 \] ### Step 2: Determine the Median The median is the middle value when the numbers are arranged in ascending order. First, sort the data: \[ 3, 4, 6, 7, 7, 7, 8, 12 \] Since there are 8 numbers (an even count), the median is the average of the 4th and 5th numbers. \[ \text{Median} = \frac{7 + 7}{2} = 7 \] ### Step 3: Identify the Mode The mode is the number that appears most frequently in the dataset. In this set, the number **7** appears **three times**, which is more frequent than any other number. \[ \text{Mode} = 7 \] ### Summary for Problem a: - **Mean:** \( 6.75 \) - **Median:** \( 7 \) - **Mode:** \( 7 \) Given the answer choices: - **A. The mode(s) is/are 7.** - **B. There is no mode.** - **b. The mean is \( \square \). (Round to the nearest thousandth as needed.)** The correct selections are: - **A. The mode(s) is/are 7.** - **b. The mean is \( 6.750 \).**

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Let's calculate the mean, median, and mode for each of the given data sets one by one. **a. Data set:** \( 3, 7, 7, 8, 6, 7, 12, 4 \) - **Mean:** \[ \text{Mean} = \frac{3 + 7 + 7 + 8 + 6 + 7 + 12 + 4}{8} = \frac{54}{8} = 6.75 \] - **Median:** First, arrange the data: \( 3, 4, 6, 7, 7, 7, 8, 12 \) Since there are 8 numbers, the median is the average of the 4th and 5th values: \[ \text{Median} = \frac{7 + 7}{2} = 7 \] - **Mode:** The mode is the most frequent number: 7 appears 3 times. Thus, for choice A, the mode(s) is/are \( 7 \). --- **b. Data set:** \( 10.5, 12.5, 12.5, 14.5, 19.5, 16.5, 12.5, 14.5, 11.5 \) - **Mean:** \[ \text{Mean} = \frac{10.5 + 12.5 + 12.5 + 14.5 + 19.5 + 16.5 + 12.5 + 14.5 + 11.5}{9} = \frac{ 12.5 \cdot 4 + 10.5 + 14.5 + 19.5 + 16.5 + 11.5 }{9} \] - **Median:** First, arrange the data: \( 10.5, 11.5, 12.5, 12.5, 12.5, 14.5, 14.5, 16.5, 19.5 \) The median is the 5th value: \[ \text{Median} = 12.5 \] - **Mode:** The mode is 12.5 (occurs 4 times). To summarize: - For data set a, the mode is 7. - For data set b, the mean is approximately \( 14.308 \). Feel free to ask if you'd like me to calculate means, medians, and modes for the other data sets as well!

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