Use formulas (5) and (6) or (5) and (7) to solve Problems...

Use formulas (5) and (6) or (5) and (7) to solve Problems 22, 23, 24, and 25. To use formulas (5) and (6) to evaluate the sample mean and standard deviation, use the following column heads: $$\begin{array}{l|l|l|l|l} \text { Midpoint } x & \text { Frequency } f & x f & (x-\bar{x}) & (x-\bar{x})^{2} & (x-\bar{x})^{2} f \\ \hline \end{array}$$ For formulas (5) and (7), use these column heads: $$\begin{array}{l|l|l|l|l} \text { Midpoint } x & \text { Frequency } f & x f & x^{2} & x^{2} f \\ \hline \end{array}$$ What was the age distribution of prehistoric Native Americans? Extensive anthropologic studies in the southwestern United States gave the following information about a prehistoric extended family group of 80 members on what is now the Navajo Reservation in northwestern New Mexico (Source: Based on information taken from Prehistory in the Navajo Reservation District, by F. W. Eddy, Museum of New Mexico Press). $$\begin{array}{l|cccc} \text { Age range (years) } & 1-10 & 11-20 & 21-30 & 31 \text { and over } \\ \hline \text { Number of individuals } & 34 & 18 & 17 & 11 \\ \hline \end{array}$$ For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint.

Use formulas (5) and (6) or (5) and (7) to solve Problems 22, 23, 24, and 25. To use formulas (5)
and (6) to evaluate the sample mean and standard deviation, use the following column heads:
$$\begin{array}{l|l|l|l|l}
\text { Midpoint } x & \text { Frequency } f & x f & (x-\bar{x}) & (x-\bar{x})^{2} & (x-\bar{x})^{2} f \\
\hline
\end{array}$$
For formulas (5) and (7), use these column heads:
$$\begin{array}{l|l|l|l|l}
\text { Midpoint } x & \text { Frequency } f & x f & x^{2} & x^{2} f \\
\hline
\end{array}$$
What was the age distribution of prehistoric Native Americans? Extensive anthropologic studies in the southwestern United States gave the following information about a prehistoric extended family group of 80 members on what is now the Navajo Reservation in northwestern New Mexico (Source: Based on information taken from Prehistory in the Navajo Reservation District, by F. W. Eddy, Museum of New Mexico Press).
$$\begin{array}{l|cccc}
\text { Age range (years) } & 1-10 & 11-20 & 21-30 & 31 \text { and over } \\
\hline \text { Number of individuals } & 34 & 18 & 17 & 11 \\
\hline
\end{array}$$
For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the class 31 and over, use 35.5 as the class midpoint.

Key Concepts

Grouped Data Analysis

Grouped data analysis involves summarizing data that has been organized into groups or classes, rather than individual data points. In statistics, this method is useful for dealing with large datasets or data that naturally falls into intervals, requiring approximation methods to estimate central tendencies and variability.

Class Midpoint

A class midpoint is the value that represents the center of a class interval in a grouped frequency distribution. It is obtained by averaging the upper and lower boundaries of the class and is used to estimate the value of each observation within that class for the purpose of calculating statistical measures.

Sample Mean

The sample mean is the arithmetic average of the data values in a sample. When dealing with grouped data, the sample mean is estimated by taking a weighted average of the class midpoints, with the weights being the frequencies of the corresponding classes. This provides an estimate of the central tendency of the dataset.

Sample Variance

The sample variance measures the spread of the data around the sample mean and is calculated as the average of the squared differences from the mean. For grouped data, it is estimated by summing the products of the frequency of each class and the squared difference between the class midpoint and the sample mean, then dividing by the degrees of freedom (typically n - 1, where n is the total frequency).

Sample Standard Deviation

The sample standard deviation is the square root of the sample variance. It provides a measure of the average distance of the data values from the sample mean, expressed in the same units as the data. When calculated for grouped data, it quantifies the spread or dispersion in the dataset and is critical for understanding the variability among the data.

Transcript

00:01 For today's question, we will be using formulas 5 and 6 from this section of the textbook.

00:06 Since we are using these two formulas, as stated in the question, we'll be using this chart to solve for our answer.

00:13 To find the midpoints, what we do is we take the lower bound of a class, and we add the upper bound of the class, and we divide by 2.

00:20 So for the first class of this data, our midpoint is 5 .5.

00:25 Doing the same thing for the rest of the classes, we get 15 .5, 25 .5.

00:30 And 35 .5, noting that the 35 .5 comes directly from the question.

00:38 And now the frequency is just the number of data in each class.

00:42 And so we had 34, 18, 17, and 11, again, coming straight from the question.

00:50 Now for this column, we just multiply the numbers from this column with this column, getting us 187, 279, 433 .5, and 390 .4 .5.

01:06 Now for this column, we need to find the sample mean for the frequency deviation.

01:13 So what we do is we take the sum of, we should be a sum, we take the sum of all of the data in this column here, and then we divide that by the number of data, and the total number of data will be the sum of all of the different frequencies.

01:36 So the sum of all of these numbers is 1290, 1 ,290, and the sum of these numbers is 80.

01:52 So we do 1 ,290 divided by 80, and we get 16 .125.

01:57 So this is our sample means...

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