To compute the sample standard deviation using the defining formula, we follow these steps:

Data Set:

7, 8, 9, 10, 11, 12, 13, 14, 155

Step 1: Compute the Mean (Average)

The sample mean ($\bar{x}$) is calculated as the sum of all the values divided by the number of values in the data set.

$\bar{x} = \frac{7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 155}{9}$

$\bar{x} = \frac{239}{9} = 26.56$

Step 2: Calculate the Squared Deviations

Now, compute the squared deviation of each data point from the mean:

$(7 - 26.56)^2 = (-19.56)^2 = 382.02$ $(8 - 26.56)^2 = (-18.56)^2 = 344.73$ $(9 - 26.56)^2 = (-17.56)^2 = 308.47$ $(10 - 26.56)^2 = (-16.56)^2 = 274.47$ $(11 - 26.56)^2 = (-15.56)^2 = 242.73$ $(12 - 26.56)^2 = (-14.56)^2 = 213.34$ $(13 - 26.56)^2 = (-13.56)^2 = 186.30$ $(14 - 26.56)^2 = (-12.56)^2 = 161.61$ $(155 - 26.56)^2 = (128.44)^2 = 16496.83$

Step 3: Compute the Sample Variance

The sample variance is calculated as the sum of the squared deviations divided by $n-1$, where $n$ is the number of data points.

$\text{Sample Variance} = \frac{382.02 + 344.73 + 308.47 + 274.47 + 242.73 + 213.34 + 186.30 + 161.61 + 16496.83}{9 - 1}$ $\text{Sample Variance} = \frac{1875.26 + 16496.83}{8} = \frac{18372.09}{8} = 2296.51$

Step 4: Compute the Sample Standard Deviation

The sample standard deviation ($s$) is the square root of the sample variance.

$s = \sqrt{2296.51} = 47.91$

Conclusion:

The sample standard deviation for the data set is approximately 47.91.

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Would you like more details on any step, or do you have any questions?

Here are 5 related questions to extend your understanding:

1. What is the difference between sample and population standard deviation?
2. How would the computation change if we had outliers in the data set?
3. What is the significance of the squared deviations in the formula?
4. How do you interpret a large standard deviation value?
5. What other measures of variability can be used besides standard deviation?

Tip: When dealing with data that includes outliers, consider whether the outliers should be excluded or weighted differently, as they can heavily impact the standard deviation.