Garza Davies
02/16/2023 · Junior High School

Find the range, mean, and standard deviation of the data set. \( \{2,11,2,15,15,10,17,6\} \) b The range is 15. (Type a whole number.) The mean is (Type an integer orsa decimal.)

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To find the range, mean, and standard deviation of the data set \( \{2, 11, 2, 15, 15, 10, 17, 6\} \), we can follow these steps: ### Step 1: Calculate the Range The range is calculated as the difference between the maximum and minimum values in the data set. - Maximum value: \( 17 \) - Minimum value: \( 2 \) \[ \text{Range} = \text{Maximum} - \text{Minimum} = 17 - 2 = 15 \] ### Step 2: Calculate the Mean The mean is calculated by summing all the values and dividing by the number of values. \[ \text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}} \] Calculating the sum: \[ 2 + 11 + 2 + 15 + 15 + 10 + 17 + 6 = 68 \] The number of values is \( 8 \). \[ \text{Mean} = \frac{68}{8} = 8.5 \] ### Step 3: Calculate the Standard Deviation To calculate the standard deviation, we first find the variance. 1. Find the mean (which we already calculated as \( 8.5 \)). 2. Calculate the squared differences from the mean: \[ (2 - 8.5)^2 = (-6.5)^2 = 42.25 \] \[ (11 - 8.5)^2 = (2.5)^2 = 6.25 \] \[ (2 - 8.5)^2 = (-6.5)^2 = 42.25 \] \[ (15 - 8.5)^2 = (6.5)^2 = 42.25 \] \[ (15 - 8.5)^2 = (6.5)^2 = 42.25 \] \[ (10 - 8.5)^2 = (1.5)^2 = 2.25 \] \[ (17 - 8.5)^2 = (8.5)^2 = 72.25 \] \[ (6 - 8.5)^2 = (-2.5)^2 = 6.25 \] 3. Sum the squared differences: \[ 42.25 + 6.25 + 42.25 + 42.25 + 42.25 + 2.25 + 72.25 + 6.25 = 213.5 \] 4. Divide by the number of values (for population standard deviation) or by \( n-1 \) (for sample standard deviation). Here, we will use \( n-1 \) since it's common to treat data sets as samples: \[ \text{Variance} = \frac{213.5}{8 - 1} = \frac{213.5}{7} \approx 30.5 \] 5. Take the square root of the variance to find the standard deviation: \[ \text{Standard Deviation} \approx \sqrt{30.5} \approx 5.52 \] ### Summary of Results - **Range**: \( 15 \) - **Mean**: \( 8.5 \) - **Standard Deviation**: \( \approx 5.52 \) Thus, the final answers are: - The range is \( 15 \). - The mean is \( 8.5 \).

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The range is 15. The mean is 8.5. The standard deviation is approximately 5.52.
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