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Calculating the mean is a fundamental aspect of descriptive statistics, providing an average that represents a data set. To find the mean age of the community of prehistoric Native Americans, you'll follow a few clear steps.

First, it's essential to determine each class midpoint. The midpoint represents the center of a class interval, indicating the most likely value for any member of that class. For example, the midpoint for ages 1-10 is calculated as: \[ \frac{1 + 10}{2} = 5.5 \] Subsequent midpoints include: \[ 11-20: \frac{11 + 20}{2} = 15.5, \quad 21-30: \frac{21 + 30}{2} = 25.5, \quad 31+ : 35.5 \] Next, multiply each midpoint by the number of individuals in the respective age range. Summing these products gives the total age sum of the group: \[ 5.5 \times 34 + 15.5 \times 18 + 25.5 \times 17 + 35.5 \times 11 = 1290.0 \] Finally, divide by the total population to find the mean age: \[ \bar{x} = \frac{1290.0}{80} = 16.125 \text{ years} \] Thus, the mean age reflects the central tendency of this community's age distribution.

Variance and standard deviation are measures of how much ages deviate from the mean. They indicate the spread or variability of the data set.The calculation starts with determining each group's deviation, which is the difference from the mean age of 16.125 years. Squaring these deviations avoids negative values and emphasizes larger differences.

• For ages 1-10, the deviation squared is \((5.5 - 16.125)^2\)
• For ages 11-20: \((15.5 - 16.125)^2\)
• For ages 21-30: \((25.5 - 16.125)^2\)
• For ages 31+: \((35.5 - 16.125)^2\)

Each squared deviation multiplies by the number of individuals in that group. Summing these results gives:\[ 4020.875 + 6.75 + 1368.875 + 4102.875 = 9499.375 \]Finally, divide by the number of samples minus one (79, in this case), yielding:\[ s^2 = \frac{9499.375}{79} \approx 120.247 \text{ years}^2 \]Taking the square root of the variance provides the standard deviation:\[ s = \sqrt{120.247} \approx 10.966 \]This result tells us that, on average, individual ages deviate from the mean by about 11 years. Variance and standard deviation offer valuable insights into the data's diversity.

Class midpoints are central in determining the average characteristics of a data set when ranges are involved. They are particularly helpful in grouped data, highlighting the average value within each class interval.To find a midpoint, calculate the average of the class endpoints. It acts like picking the middle point of a range, assuming all within that range are equally likely.Consider these calculations:

• Ages 1-10 midpoint: \( \frac{1 + 10}{2} = 5.5 \)
• Ages 11-20 midpoint: \( \frac{11 + 20}{2} = 15.5 \)
• Ages 21-30 midpoint: \( \frac{21 + 30}{2} = 25.5 \)
• Ages 31+: The midpoint is already assigned as 35.5.

Each class midpoint represents an assumed center for all individuals in its range. When estimating characteristics like the mean age, midpoints simplify complex ranges into single representative values. In practice, they are vital in organizing and summarizing survey results or data sets efficiently.