Q1 Frequency Distribution of Body T... [FREE SOLUTION]

Chapter 2: Q1 (page 76)

Frequency Distribution of Body Temperatures Construct a frequency distribution of the 20 body temperatures (oF) listed below. (These data are from Data Set 3 鈥楤ody Temperatures鈥� in Appendix B.) Use a class width of 0.5oF and a starting value of 97.0oF.
97.1 97.2 97.5 97.6 97.6 97.8 98.0 98.0 98.2 98.2 98.2 98.3 98.4 98.6 98.6 98.7 98.7 98.9 99.1 99.4.

Short Answer

Expert verified

The following frequency distribution is constructed for the given data on body temperatures.

Body Temperature	Frequency
97.0-97.4	2
97.5-97.9	4
98.0-98.4	7
98.5-98.9	5
99.0-99.4	2

Step by step solution

Given information

Data on the body temperatures of 20 subjects measured in degrees Fahrenheit are given.

Define a frequency distribution

A frequency distribution is an arrangement of data values in the form of closed intervals.

The frequency of each class interval is tabulated by counting the number of data values that fall in each interval. A grouped frequency distribution can be continuous or not.

Describe the lower limits of the class intervals

An inclusive type of frequency distribution is constructed where the class width is equal to 0.5 degrees Fahrenheit, and the first value is equal to 97.0 degrees Fahrenheit.

The maximum value in the data is equal to 99.4 degrees Fahrenheit.

The lower limit of the first class interval is equal to 97.0 degrees Fahrenheit.

The class width is 0.5 degrees Fahrenheit.

According to the given formula, the lower class limits of the five intervals are computed as below.

$Class width = L . L_{i} - L L_{i - 1}$ for $L L_{i}, L L_{i + 1}$ as the lower limits of the (i)th and (i+1)th class intervals.

Thus, the lower limits for each of the classes are as follows.

$L L_{1} = 97.0 L L_{2} = 97.0 + 0.5 = 97.5$

$L L_{3} = 97.5 + 0.5 = 98.0$

$L L_{4} = 98.0 + 0.5 = 98.5 L L_{5} = 98.5 + 0.5 = 99.0$

$L L_{6} = 99.0 + 0.5 = 99.5$

Describe the upper limits of the class intervals

Considering a gap of 1 unit in between each successive interval (as the values are rounded up to one decimal place),the following upper-class limits are constructed.

$U L_{1} = L L_{2} - 0.1 = 97.5 - 0.1 = 97.4$

$U L_{2} = L L_{3} - 0.1 = 98.0 - 0.1 = 97.9$

$U L_{3} = L L_{4} - 0.1 = 98.5 - 0.1 = 98.4$

$U L_{4} = L L_{5} - 0.1 = 99.0 - 0.1 = 98.9$

$U L_{5} = L L_{6} - 0.1 = 99.5 - 0.1 = 99.4$

Find the frequency distribution of the observations

Describe the upper and lower limits as class intervals, and obtain the corresponding counts of temperatures that lie in each interval (both limits are inclusive).

Thus, the following frequency distribution is constructed.