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Let's calculate the sample variance and standard deviation for the dataset: 22, 10, 4, 9, 11.

### Step 1: Find the mean (\(\bar{x}\))
\[
\bar{x} = \frac{22 + 10 + 4 + 9 + 11}{5} = \frac{56}{5} = 11.2
\]

### Step 2: Calculate the squared differences from the mean for each value
\[
\begin{align*}
(22 - 11.2)^2 &= (10.8)^2 = 116.64 \\
(10 - 11.2)^2 &= (-1.2)^2 = 1.44 \\
(4 - 11.2)^2 &= (-7.2)^2 = 51.84 \\
(9 - 11.2)^2 &= (-2.2)^2 = 4.84 \\
(11 - 11.2)^2 &= (-0.2)^2 = 0.04
\end{align*}
\]

### Step 3: Find the sum of squared differences
\[
116.64 + 1.44 + 51.84 + 4.84 + 0.04 = 174.8
\]

### Step 4: Calculate the sample variance (\(s^2\))
Since this is a sample, we divide by \(n - 1\) (degrees of freedom), where \(n = 5\):
\[
s^2 = \frac{174.8}{5 - 1} = \frac{174.8}{4} = 43.7
\]

### Step 5: Calculate the sample standard deviation (\(s\))
\[
s = \sqrt{43.7} \approx 6.6
\]

Thus:
- The sample variance \(s^2 = 43.7\)
- The sample standard deviation \(s \approx 6.6\)

Would you like further details or have any questions?

Here are 5 related questions for more practice:
1. What is the population variance for this dataset if it were considered a population?
2. How would the variance change if we added a new number to the dataset?
3. Can you compute the variance and standard deviation for a different dataset, such as 15, 20, 25, 30, 35?
4. How does the variance relate to the spread of the data?
5. What is the difference between sample variance and population variance?

**Tip:** The sample variance uses \(n - 1\) in the denominator to account for bias in small samples.

Find the sample variance and standard deviation. Dataset: 22, 10, 4, 9, 11.

Learn how to calculate the sample variance and standard deviation for the dataset 22, 10, 4, 9, 11. Step-by-step calculations including finding the mean, sum of squared differences, and final variance and standard deviation values. Suitable for students in Grades 9-12.

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