QuestionAugust 25, 2025

What is the standard deviation of this data set? 0,1,2,2,3,3,3,4,4,5,6

What is the standard deviation of this data set? 0,1,2,2,3,3,3,4,4,5,6
What is the standard deviation of this data set? 0,1,2,2,3,3,3,4,4,5,6

Solution
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Answer

\sqrt{\frac{30}{11}} \approx 1.65 Explanation 1. Calculate the Mean The mean \bar{x} is calculated as \bar{x} = \frac{\sum x_i}{n} = \frac{0+1+2+2+3+3+3+4+4+5+6}{11} = \frac{33}{11} = 3. 2. Calculate the Variance Variance \sigma^2 is \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(0-3)^2 + (1-3)^2 + (2-3)^2 + (2-3)^2 + (3-3)^2 + (3-3)^2 + (3-3)^2 + (4-3)^2 + (4-3)^2 + (5-3)^2 + (6-3)^2}{11} = \frac{30}{11}. 3. Calculate the Standard Deviation Standard deviation \sigma is \sigma = \sqrt{\sigma^2} = \sqrt{\frac{30}{11}}.

Explanation

1. Calculate the Mean<br /> The mean $\bar{x}$ is calculated as $\bar{x} = \frac{\sum x_i}{n} = \frac{0+1+2+2+3+3+3+4+4+5+6}{11} = \frac{33}{11} = 3$.<br />2. Calculate the Variance<br /> Variance $\sigma^2$ is $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(0-3)^2 + (1-3)^2 + (2-3)^2 + (2-3)^2 + (3-3)^2 + (3-3)^2 + (3-3)^2 + (4-3)^2 + (4-3)^2 + (5-3)^2 + (6-3)^2}{11} = \frac{30}{11}$.<br />3. Calculate the Standard Deviation<br /> Standard deviation $\sigma$ is $\sigma = \sqrt{\sigma^2} = \sqrt{\frac{30}{11}}$.
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