To calculate the sample standard deviation, we can follow these steps:

Step 1: Find the mean ($\bar{x}$) of the data.

$\bar{x} = \frac{\sum x_i}{n} = \frac{19 + 25 + 13 + 28 + 20 + 11 + 22 + 14}{8} = \frac{152}{8} = 19$

Step 2: Subtract the mean from each data point and square the result.

$(x_i - \bar{x})^2$ $(19 - 19)^2 = 0^2 = 0$ $(25 - 19)^2 = 6^2 = 36$ $(13 - 19)^2 = (-6)^2 = 36$ $(28 - 19)^2 = 9^2 = 81$ $(20 - 19)^2 = 1^2 = 1$ $(11 - 19)^2 = (-8)^2 = 64$ $(22 - 19)^2 = 3^2 = 9$ $(14 - 19)^2 = (-5)^2 = 25$

Step 3: Sum all the squared differences.

$\sum (x_i - \bar{x})^2 = 0 + 36 + 36 + 81 + 1 + 64 + 9 + 25 = 252$

Step 4: Divide by the number of data points minus one ($n - 1$) to find the variance.

$\text{Variance} = \frac{\sum (x_i - \bar{x})^2}{n - 1} = \frac{252}{8 - 1} = \frac{252}{7} = 36$

Step 5: Take the square root of the variance to find the sample standard deviation.

$\text{Sample Standard Deviation} = \sqrt{36} = 6.00$

So, the sample standard deviation is 6.00 (rounded to two decimal places).

Do you have any questions or need more details?

Here are 5 related questions:

1. How does the sample standard deviation differ from the population standard deviation?
2. What would happen to the standard deviation if one of the data points was much larger or smaller?
3. How is standard deviation useful in understanding the spread of a dataset?
4. Can you explain the difference between variance and standard deviation?
5. What are some examples of when a high standard deviation might be expected in real-world data?

Tip: Remember that the sample standard deviation is a measure of how spread out the data is around the mean. The larger the standard deviation, the more spread out the data points are.