To calculate the standard deviation of the dataset $9.1, 9.4, 10, 9.5, 7.7$, we follow these steps:

Step 1: Find the Mean ($\mu$)

The mean is calculated as: $\mu = \frac{\text{Sum of all data points}}{\text{Number of data points}}$ $\mu = \frac{9.1 + 9.4 + 10 + 9.5 + 7.7}{5} = \frac{45.7}{5} = 9.14$

Step 2: Calculate the Variance ($\sigma^2$)

The variance is the average of the squared differences between each data point and the mean. For each data point $x_i$: $\text{Variance} (\sigma^2) = \frac{\sum_{i=1}^n (x_i - \mu)^2}{n}$

1. $(9.1 - 9.14)^2 = (-0.04)^2 = 0.0016$
2. $(9.4 - 9.14)^2 = (0.26)^2 = 0.0676$
3. $(10 - 9.14)^2 = (0.86)^2 = 0.7396$
4. $(9.5 - 9.14)^2 = (0.36)^2 = 0.1296$
5. $(7.7 - 9.14)^2 = (-1.44)^2 = 2.0736$

Sum of squared differences: $0.0016 + 0.0676 + 0.7396 + 0.1296 + 2.0736 = 3.011$

Variance: $\sigma^2 = \frac{3.011}{5} = 0.6022$

Step 3: Calculate the Standard Deviation ($\sigma$)

The standard deviation is the square root of the variance: $\sigma = \sqrt{0.6022} \approx 0.776$

Final Answer:

The standard deviation of the dataset is approximately 0.776.

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Would you like a detailed explanation of any step?
Here are 5 related questions for further exploration:

1. What is the difference between population and sample standard deviation?
2. How is standard deviation used in real-world data analysis?
3. Can you explain why we square the differences in variance calculations?
4. How would the standard deviation change if an outlier is added to the dataset?
5. What is the significance of a small vs. large standard deviation?

Tip: Always double-check whether you're calculating for a population or a sample, as the formula slightly differs.