Davison John
08/20/2023 · Senior High School

he length (in pages) of math research projects is given elow. Using this information, calculate the range, ariance, and standard deviation. \[ 19,16,34,11,29,24,16,299,21 \] ange \( = \) ariance \( = \) tandard deviation =

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To calculate the range, variance, and standard deviation of the given data set, we will follow these steps: 1. **Calculate the Range**: The range is the difference between the maximum and minimum values in the data set. Given data: \( 19, 16, 34, 11, 29, 24, 16, 299, 21 \) - Maximum value = 299 - Minimum value = 11 \[ \text{Range} = \text{Maximum} - \text{Minimum} = 299 - 11 = 288 \] 2. **Calculate the Variance**: Variance is calculated using the formula: \[ \text{Variance} = \frac{\sum (x_i - \mu)^2}{N} \] where \( \mu \) is the mean of the data set, \( x_i \) are the individual data points, and \( N \) is the number of data points. - First, calculate the mean (\( \mu \)): \[ \mu = \frac{\sum x_i}{N} = \frac{19 + 16 + 34 + 11 + 29 + 24 + 16 + 299 + 21}{9} = \frac{ 19 + 16 + 34 + 11 + 29 + 24 + 16 + 299 + 21 }{9} = \frac{ 449 }{9} \approx 49.89 \] - Next, calculate the squared differences from the mean: \[ \begin{align*} (19 - 49.89)^2 & \approx 952.5921 \\ (16 - 49.89)^2 & \approx 1135.5921 \\ (34 - 49.89)^2 & \approx 252.5921 \\ (11 - 49.89)^2 & \approx 1495.5921 \\ (29 - 49.89)^2 & \approx 436.5921 \\ (24 - 49.89)^2 & \approx 653.5921 \\ (16 - 49.89)^2 & \approx 1135.5921 \\ (299 - 49.89)^2 & \approx 61756.5921 \\ (21 - 49.89)^2 & \approx 826.5921 \\ \end{align*} \] - Now sum these squared differences: \[ \sum (x_i - \mu)^2 \approx 952.5921 + 1135.5921 + 252.5921 + 1495.5921 + 436.5921 + 653.5921 + 1135.5921 + 61756.5921 + 826.5921 \approx 65192.0 \] - Finally, calculate the variance: \[ \text{Variance} = \frac{65192.0}{9} \approx 7243.56 \] 3. **Calculate the Standard Deviation**: The standard deviation is the square root of the variance. \[ \text{Standard Deviation} = \sqrt{\text{Variance}} \approx \sqrt{7243.56} \approx 85.0 \] ### Summary of Results: - Range \( = 288 \) - Variance \( \approx 7243.56 \) - Standard Deviation \( \approx 85.0 \) Please note that the calculations for variance and standard deviation are based on the assumption that the data represents a population. If it represents a sample, the denominator in the variance calculation would be \( N-1 \) instead of \( N \).

Quick Answer

Range = 288 Variance ≈ 7243.56 Standard Deviation ≈ 85.0
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