Question

In Exercises 5–8, identify the class width, class midpoints, and class boundaries for the given frequency distribution. Also identify the number of individuals included in the summary. The frequency distributions are based on real data from Appendix B.

Blood Platelet Count of Females	Frequency
100-199	25
200-299	92
300-399	28
400-499	0
500-599	2

Short answer

The class width for each interval is equal to 100.

The midpoints are 149.5, 249.5, 349.5, 449.5, and 549.5.

The class boundaries are 99.5, 199.5, 299.5, 399.5, 499.5, and 599.5.

The total number of individuals included is equal to 147.

Step-by-step solution

Given information

Data are given on the blood platelet counts of males.

Class width

The class width of an interval is computed by subtracting two consecutive lower class limits.

Here, the lower class limit of the first class interval is equal to 100, and the lower class limit of the second class interval is equal to 200.

Thus, the class width is as follows.

$$\begin{gathered} \text{Class}\text{width}=200-100 \\ =100 \end{gathered}$$

As it can be observed that all class intervals have equal width, the class width for each interval is equal to 100.

Class midpoint

The midpoint of a class interval is calculated using the following formula:

$\text{Midpoint}=\frac{\text{lower}\text{limit}+\text{upper}\text{limit}}{2}$

Thus, the midpoints of the class intervals are computed as shown below.

Class interval	Midpoint
100-199	$$\begin{gathered} \text{Midpoint}=\frac{100+199}{2} \\ =149.5 \end{gathered}$$
200-299	$$\begin{gathered} \text{Midpoint}=\frac{200+299}{2} \\ =249.5 \end{gathered}$$
300-399	$$\begin{gathered} \text{Midpoint}=\frac{300+399}{2} \\ =349.5 \end{gathered}$$
400-499	$$\begin{gathered} \text{Midpoint}=\frac{400+499}{2} \\ =449.5 \end{gathered}$$
500-599	$$\begin{gathered} \text{Midpoint}=\frac{500+599}{2} \\ =549.5 \end{gathered}$$

Thus, the midpoints are 149.5, 249.5, 349.5, 449.5, and 549.5.

Class boundaries

The value of the gap between each successive interval divided by 2 is subtracted from the lower limit and added to the upper limit of a class interval to obtain the class boundaries.

The gap is calculated as shown below.

$$\begin{gathered} \text{Gap}=2^{nd}\text{lower}\text{class}\text{limit}-1^{st}\text{upper}\text{class}\text{limit} \\ =100-99 \\ =1 \end{gathered}$$

The value equal to is subtracted from the lower class limits and added to the upper-class limits of each interval. Thus, the class boundaries are obtained as shownbelow.

Class interval	Lower class boundaries	Upper class boundaries
100-199	$100-0.5=99.5$	$199+0.5=199.5$
200-299	$200-0.5=199.5$	$299+0.5=299.5$
300-399	$300-0.5=299.5$	$399+0.5=399.5$
400-499	$400-0.5=399.5$	$499+0.5=499.5$
500-599	$500-0.5=499.5$	$599+0.5=599.5$

Therefore, the class boundaries are 99.5, 199.5, 299.5, 399.5, 499.5, and 599.5.

Total number of individuals

The total number of individuals included in the given frequency distribution is equal to the sum of the frequencies of all classes. The total frequency is computed below.

$$\begin{gathered} \text{Total}\text{frequency}=25+92+28+0+2 \\ =147 \end{gathered}$$

Therefore, the total number of individuals included in the given frequency distribution is 147.