Let's solve the problem step-by-step based on the image provided. The given sample data set is:

$x = \{26.8, 5.7, 8.6, 21, 24.1, 23.6, 27.1\}$

Part (a) - Finding the Mean:

The formula for the mean (average) of a data set is:

$\text{Mean} = \frac{\sum{x_i}}{n}$

where $n$ is the number of data points.

1. Sum of the dataLet's solve the problem step-by-step based on the image provided. The given sample data set is:

$x = \{26.8, 5.7, 8.6, 21, 24.1, 23.6, 27.1\}$

Part (a) - Finding the Mean:

The formula for the mean (average) of a data set is:

$\text{Mean} = \frac{\sum{x_i}}{n}$

where $n$ is the number of data points.

1. Sum of the $26.8 + 5.7 + 8.6 + 21 + 24.1 + 23.6 + 27.1 = 136.9$

2. Number of data points: $n = 7$

3. Calculate the mean: $\text{Mean} = \frac{136.9}{7} = 19.56$

So, the mean of the data is: $\boxed{19.56}$

Part (b) - Finding the Sample Standard Deviation:

The formula for the sample standard deviation is:

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

where:

• $x_i$ are the data points,
• $\bar{x}$ is the mean (19.56),
• $n$ is the number of data points (7).

Step 1: Calculate the squared differences from the mean.

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

• $(26.8 - 19.56)^2 = 52.49$
• $(5.7 - 19.56)^2 = 192.34$
• $(8.6 - 19.56)^2 = 120.34$
• $(21 - 19.56)^2 = 2.07$
• $(24.1 - 19.56)^2 = 20.60$
• $(23.6 - 19.56)^2 = 16.39$
• $(27.1 - 19.56)^2 = 56.71$

Step 2: Sum the squared differences.

$52.49 + 192.34 + 120.34 + 2.07 + 20.60 + 16.39 + 56.71 = 460.94$

Step 3: Divide by $n - 1 = 6$.

$\frac{460.94}{6} = 76.82$

Step 4: Take the square root of the result.

$s = \sqrt{76.82} = 8.76$

So, the sample standard deviation is: $\boxed{8.76}$

Final Answers:

• Mean: $19.56$
• Sample Standard Deviation: $8.76$

Do you have any questions or need further clarification?

5 Related Questions:

1. What is the difference between sample and population standard deviation?
2. How would the mean change if a new data point (e.g., 10) was added?
3. How can you calculate variance from the standard deviation?
4. What are some real-life examples of using standard deviation?
5. How does removing an outlier affect the standard deviation?

Tip:

Always use the sample standard deviation formula when working with smaller datasets, as it adjusts for the fact that you are not measuring the entire population.