The ages of the sti ien volunteers at an event are: \( 13,11,12,15,16,11,12,14,15,11 \)

1. **List the ages:** The given ages are: \[ 13,\ 11,\ 12,\ 15,\ 16,\ 11,\ 12,\ 14,\ 15,\ 11 \] 2. **Compute the Mean (Average):** - First, compute the sum of the ages: \[ 13+11+12+15+16+11+12+14+15+11 = 130 \] - There are \( n = 10 \) ages. - The mean is: \[ \text{Mean} = \frac{130}{10} = 13 \] 3. **Determine the Median:** - Sort the ages in increasing order: \[ 11,\ 11,\ 11,\ 12,\ 12,\ 13,\ 14,\ 15,\ 15,\ 16 \] - With \( n = 10 \) (an even number), the median is the average of the 5th and 6th numbers. The 5th number is \( 12 \) and the 6th number is \( 13 \). - Thus, the median is: \[ \text{Median} = \frac{12+13}{2} = \frac{25}{2} = 12.5 \] 4. **Find the Mode:** - Count the frequency of each age: - \( 11 \) appears 3 times - \( 12 \) appears 2 times - \( 15 \) appears 2 times - Others appear once - The age that occurs most frequently is \( 11 \), so the mode is: \[ \text{Mode} = 11 \] 5. **Calculate the Range:** - The minimum age is \( 11 \) and the maximum age is \( 16 \). - The range is: \[ \text{Range} = 16 - 11 = 5 \] 6. **Compute the Variance and Standard Deviation (Population):** - The mean is \( 13 \). Compute the squared difference for each age: \[ \begin{array}{c|c} \text{Age} & (\text{Age} - 13)^2 \\ \hline 13 & (13-13)^2 = 0 \\ 11 & (11-13)^2 = 4 \\ 12 & (12-13)^2 = 1 \\ 15 & (15-13)^2 = 4 \\ 16 & (16-13)^2 = 9 \\ 11 & (11-13)^2 = 4 \\ 12 & (12-13)^2 = 1 \\ 14 & (14-13)^2 = 1 \\ 15 & (15-13)^2 = 4 \\ 11 & (11-13)^2 = 4 \\ \end{array} \] - Sum of squared differences: \[ 0+4+1+4+9+4+1+1+4+4 = 32 \] - The population variance is: \[ \text{Variance} = \frac{32}{10} = 3.2 \] - The population standard deviation is the square root of the variance: \[ \text{Standard Deviation} = \sqrt{3.2} \approx 1.79 \] Thus, the summary of the computed statistics is: - Mean: \( 13 \) - Median: \( 12.5 \) - Mode: \( 11 \) - Range: \( 5 \) - Variance: \( 3.2 \) - Standard Deviation: \( \approx 1.79 \)