Archaeology: New Mexico The Wind Mountain excavation site

Step 1 of 5

A data set has values ranging from a low of 10 to a high of 50. The class width is to be 10. What's wrong with using the class 10-20, 21-31, 32-42, 43-53 for a frequency distribution.

(a) From the known data set, the largest value is 200, the  lowest value is 10, and the number of classes is 7.

\(\begin{aligned} \text { Class width } & =\frac{\text { Largest data value }- \text { Smallest data value }}{\text { Desired no. class }} \\ & =\frac{200-10}{7} \\ & =27.14 \\ & \approx 28 \end{aligned}\)

(b) The formula for the midpoint is

\(\text { Midpoint }=\frac{\text { Lower class limit }+ \text { Upper class limit }}{2}\)

To find the upper-class boundaries, add 0.5 unit to the upper-class limits.

To find the lower-class boundaries, subtract 0.5 unit to the lower-class limits.

The formula for the relative class frequency is,

\(\begin{aligned} \text { Relative class frequency } & =\frac{f}{n} \\ & =\frac{\text { class frequency }}{\text { Total of all frequencies }} \end{aligned}\)

The frequency table is as follows:

\(\begin{array}{ccccc} \hline \begin{array}{c} \text { Class Limits } \\ \text { Lower-Upper } \end{array} & \begin{array}{c} \text { Class Boundaries } \\ \end{array} & \text { Midpoint } & \begin{array}{c} \text { Frequency } \\ n \end{array} & \begin{array}{c} \text { Realitive } \\ \text { Frequency f/n } \end{array} \\ \hline 10-37 & 9.5-37.5 & 23.5 & 7 & 0.1 \\ 38-65 & 37.5-65.5 & 51.5 & 25 & 0.34 \\ 66-93 & 65.5-93.5 & 79.5 & 26 & 0.36 \\ 94-121 & 93.5-121.5 & 107.5 & 9 & 0.12 \\ 122-149 & 121.5-149.5 & 135.5 & 5 & 0.07 \\ 150-177 & 149.5-177.5 & 163.5 & 0 & 0.0 \\ 178-205 & 177.5-205.5 & 191.5 & 1 & 0.01 \\ & & & & \\ \hline \end{array}\)