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

Step 1: Find the mean (average) of the data.

The data provided is:

$x = \{2.1, 22.6, 8.4, 19.1, 27.1, 8.0, 27.7\}$

The mean $\bar{x}$ is calculated as:

$\bar{x} = \frac{\sum x_i}{n} = \frac{2.1 + 22.6 + 8.4 + 19.1 + 27.1 + 8.0 + 27.7}{7}$

Step 2: Calculate each data point's deviation from the mean and square it.

Step 3: Find the variance.

The variance $s^2$ for a sample is given by:

$s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}$

Step 4: Calculate the standard deviation.

The standard deviation $s$ is the square root of the variance:

$s = \sqrt{s^2}$

Let's calculate these values.The mean of the sample data is $16.43$.

• Variance: $104.49$
• Standard Deviation: $10.22$

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the standard deviation change if the data were a population instead of a sample?
2. What is the significance of a high standard deviation in a dataset?
3. How does variance relate to standard deviation conceptually?
4. Can you interpret the variance and standard deviation in the context of this data?
5. How would adding an outlier affect the standard deviation?

Tip: Variance is always in squared units of the original data, while standard deviation is in the same units as the data itself.