91Ó°ÊÓ

To understand the average age of the prehistoric Native American community, we calculate the mean, which represents the central tendency of a data set. In this particular exercise, the data is grouped in age ranges, thereby requiring us to determine the midpoint for each age range first.
Here's how midpoints work:

• For each age range, the midpoint is calculated by averaging the lower and upper limits.
• This creates a single value (midpoint) that represents the entire age group.

The midpoints calculated for the different age groups were:

• 1-10 years: \((1 + 10)/2 = 5.5\)
• 11-20 years: \((11 + 20)/2 = 15.5\)
• 21-30 years: \((21 + 30)/2 = 25.5\)
• 31 years and over: Provided as 35.5

The mean age, denoted as \( \bar{x} \), is calculated using the weighted mean formula, which factors in both the midpoints and their respective frequencies.

• Formula: \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \)

In this case: \( \bar{x} = \frac{(34 \times 5.5) + (18 \times 15.5) + (17 \times 25.5) + (11 \times 35.5)}{80} = 16.125 \).The mean age of the community is 16.125 years. This result gives us a clear understanding of the age distribution's center in this prehistoric community.

Sample variance provides insight into the spread and variability of age data in the community. It tells us how much the ages differ from the mean age. To compute this, we examine each age group and calculate how far it deviates from the mean age we calculated earlier.
The calculation process involves these steps:

• Subtract the mean \( \bar{x} \) from each midpoint.
• Square this result to ensure that both negative and positive deviations contribute equally.
  
• Multiply the squared result by the frequency of each age group.

This produces the squared differences weighted by frequency:

• For 1-10 years: \((5.5 - 16.125)^2 \times 34 = 3807.863\)
• For 11-20 years: \((15.5 - 16.125)^2 \times 18 = 12.38\)
• For 21-30 years: \((25.5 - 16.125)^2 \times 17 = 1399.938\)
• For 31 and over: \((35.5 - 16.125)^2 \times 11 = 4140.38\)

Finally, to find the sample variance, sum these results and divide by the total number of instances minus one:

• \( s^2 = \frac{3807.863 + 12.38 + 1399.938 + 4140.38}{79} = 118.485 \)

The variance of 118.485 tells us how much the ages in this group are spread out around the mean.

Standard deviation is a key statistical measure that describes how data spreads out from the mean. It is derived from the variance, making it an essential concept in understanding data variability within the sample.
Here's why standard deviation matters:

• It converts the squared units of variance into the same units as the original data, thus making it easier to interpret.
• A lower standard deviation indicates that the data points tend to be close to the mean, whereas a higher standard deviation indicates a wider spread of ages.

In this exercise, the standard deviation is calculated by taking the square root of the sample variance.

• Given the variance from our calculations: \( s^2 = 118.485 \).
• The standard deviation, \( s \), is: \( \sqrt{118.485} \approx 10.88 \).

Thus, the standard deviation of approximately 10.88 years indicates moderate variability around the mean age within the population. This suggests that while some individuals are close in age to the mean, others may be significantly younger or older.