Given Data:

$5, 10, 6, 9, 8, 4, 4, 8, 9, 35$

Step 1: Range

The range is the difference between the maximum and minimum values in the dataset.

• Maximum value = 35
• Minimum value = 4

$\text{Range} = \text{Maximum} - \text{Minimum} = 35 - 4 = 31$

Step 2: Population Variance

The formula for the population variance is:

$\sigma^2 = \frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2$

where:

• $N$ is the population size (in this case, 10),
• $x_i$ are the data points,
• $\mu$ is the population mean.

a. Calculate the mean ($\mu$):

$\mu = \frac{\sum x_i}{N} = \frac{5 + 10 + 6 + 9 + 8 + 4 + 4 + 8 + 9 + 35}{10} = \frac{98}{10} = 9.8$

b. Calculate the squared differences from the mean:

\begin{align*} (5 - 9.8)^2 & = 22.84 \\ (10 - 9.8)^2 & = 0.04 \\ (6 - 9.8)^2 & = 14.44 \\ (9 - 9.8)^2 & = 0.64 \\ (8 - 9.8)^2 & = 3.24 \\ (4 - 9.8)^2 & = 33.64 \\ (4 - 9.8)^2 & = 33.64 \\ (8 - 9.8)^2 & = 3.24 \\ (9 - 9.8)^2 & = 0.64 \\ (35 - 9.8)^2 & = 624.64 \end{align*}$$ #### c. Sum of the squared differences: $$\sum (x_i - \mu)^2 = 22.84 + 0.04 + 14.44 + 0.64 + 3.24 + 33.64 + 33.64 + 3.24 + 0.64 + 624.64 = 736.00$$ #### d. Calculate the population variance: $$\sigma^2 = \frac{736.00}{10} = 73.6$$ ### Step 3: **Population Standard Deviation** The standard deviation is the square root of the variance: $$\sigma = \sqrt{73.6} = 8.58$$ ### Final Results: - **Range**: 31 - **Population Variance**: 73.6 - **Population Standard Deviation**: 8.58 Would you like to ask any further questions or need more details? ### Related Questions: 1. How is sample variance different from population variance? 2. What are the implications of a high standard deviation in a data set? 3. How would the variance change if one of the extreme values was removed? 4. Can the range ever be larger than the highest data point? 5. How do you calculate the mean of grouped data? ### Tip: The range only considers the extreme values, while variance and standard deviation provide a better picture of how the data is spread around the mean.