91影视

Data are given on the blood platelet counts of males.

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.

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.

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.

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

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.