To calculate the standard deviation and variance of the sample, we follow these steps:

Step 1: Calculate the mean (average) of the sample.

$\text{Mean} (\bar{x}) = \frac{\sum x_i}{n}$ Where $\sum x_i$ is the sum of all data points, and $n$ is the number of data points.

$\bar{x} = \frac{11.2 + 14.4 + 20.4 + 15 + 11.7 + 3.5}{6} = \frac{76.2}{6} \approx 12.7$

Step 2: Calculate each deviation from the mean, then square it.

$(x_i - \bar{x})^2$

• For $x_1 = 11.2$: $(11.2 - 12.7)^2 = (-1.5)^2 = 2.25$
• For $x_2 = 14.4$: $(14.4 - 12.7)^2 = (1.7)^2 = 2.89$
• For $x_3 = 20.4$: $(20.4 - 12.7)^2 = (7.7)^2 = 59.29$
• For $x_4 = 15$: $(15 - 12.7)^2 = (2.3)^2 = 5.29$
• For $x_5 = 11.7$: $(11.7 - 12.7)^2 = (-1.0)^2 = 1.00$
• For $x_6 = 3.5$: $(3.5 - 12.7)^2 = (-9.2)^2 = 84.64$

Step 3: Calculate the variance.

$\text{Variance} (s^2) = \frac{\sum (x_i - \bar{x})^2}{n-1}$ Where $n-1$ is the degrees of freedom for the sample.

$s^2 = \frac{2.25 + 2.89 + 59.29 + 5.29 + 1.00 + 84.64}{5} = \frac{155.36}{5} = 31.07$

Step 4: Calculate the standard deviation.

$\text{Standard Deviation} (s) = \sqrt{s^2} = \sqrt{31.07} \approx 5.57$

Final Results:

• Variance: 31.07
• Standard Deviation: 5.57

Would you like any further details or explanations?

Here are 5 related questions:

1. How would you calculate the variance for a population instead of a sample?
2. What does the standard deviation tell you about the data set?
3. How would adding an outlier affect the variance and standard deviation?
4. What are other measures of data spread besides standard deviation and variance?
5. Can the standard deviation ever be negative?

Tip: Always check whether you're dealing with a sample or a population when calculating variance and standard deviation, as the formulas slightly differ.