Let's solve the problem in the image step-by-step.

Step 1: Compute the sample mean $\bar{x}$

The data provided is:
$x = [18, 16, 28, 25, 11]$

The sample mean $\bar{x}$ is calculated as: $\bar{x} = \frac{18 + 16 + 28 + 25 + 11}{5} = \frac{98}{5} = 19.6$

So, the sample mean $\bar{x}$ is 19.6.

Step 2: Compute the squared deviations from the mean

For each value of $x$, subtract the mean $\bar{x} = 19.6$, then square the result: $(x - \bar{x})^2$

• For $x = 18$: $(18 - 19.6)^2 = (-1.6)^2 = 2.56$
• For $x = 16$: $(16 - 19.6)^2 = (-3.6)^2 = 12.96$
• For $x = 28$: $(28 - 19.6)^2 = (8.4)^2 = 70.56$
• For $x = 25$: $(25 - 19.6)^2 = (5.4)^2 = 29.16$
• For $x = 11$: $(11 - 19.6)^2 = (-8.6)^2 = 73.96$

The squared deviations are: $[2.56, 12.96, 70.56, 29.16, 73.96]$

Step 3: Compute the sample variance $s^2$

The sample variance formula is: $s^2 = \frac{\sum (x - \bar{x})^2}{n - 1}$ where $n = 5$ is the sample size.

First, sum up the squared deviations: $2.56 + 12.96 + 70.56 + 29.16 + 73.96 = 189.2$

Now compute the sample variance: $s^2 = \frac{189.2}{5 - 1} = \frac{189.2}{4} = 47.3$

So, the sample variance $s^2$ is 47.3.

Step 4: Compute the sample standard deviation $s$

The sample standard deviation is the square root of the variance: $s = \sqrt{47.3} \approx 6.88$

Thus, the sample standard deviation $s$ is approximately 6.88.

Let me know if you'd like further details or have any questions!

Related Questions:

1. What is the importance of using $n - 1$ in the sample variance formula instead of $n$?
2. How does the standard deviation change if we add a very large number to the dataset?
3. Can the standard deviation ever be negative? Why or why not?
4. What is the difference between population standard deviation and sample standard deviation?
5. How does removing an outlier affect the standard deviation?

Tip:

When interpreting the standard deviation, remember that it provides insight into the variability of your data around the mean.