Exercise solution 8, page 67 - Mathematics Higher Student Book

⭐ What you need to know:

A) A grouped frequency table shows how many values fall within each interval. For example, a class such as \(30 \leq t < 50\) means the time is at least \(30\) minutes, but less than \(50\) minutes. The lower bound is included, and the upper bound is not included.

B) To find the total number of runners, add all the frequencies in the table. The frequency tells you how many data values are in each class.

For example, if the frequencies were \(6\), \(8\), \(11\) and \(5\), then

\[\begin{aligned} \text{Total} &= 6+8+11+5 \\ &= 30 \end{aligned}\]

C) To find a percentage, use

\[ \text{percentage}=\frac{\text{part}}{\text{whole}}\times 100 \]

In a grouped table, first identify which classes match the condition in the question, then add their frequencies to get the part.

For example, if \(18\) runners out of \(45\) took more than a certain time, then

\[\begin{aligned} \text{percentage} &= \frac{18}{45}\times 100 \\ &= 0.4\times 100 \\ &= 40\% \end{aligned}\]

D) The range tells you how spread out the data is. It is found using

\[ \text{range}=\text{largest value}-\text{smallest value} \]

When the data is grouped into intervals, you do not know the exact smallest and largest values. That is why the range is only an estimate. You usually use the class boundaries or the end values of the first and last intervals involved.

For example, if the first class is \(15 \leq t < 35\) and the last class is \(95 \leq t < 115\), an estimated range is

\[\begin{aligned} \text{estimated range} &= 115-15 \\ &= 100 \end{aligned}\]

This is only an estimate because the actual smallest value might be bigger than \(15\), and the actual largest value might be smaller than \(115\).

E) A frequency polygon is drawn by plotting the midpoint of each class interval against its frequency, then joining the points with straight lines.

To find a class midpoint, add the two class boundaries and divide by \(2\):

\[ \text{midpoint}=\frac{\text{lower bound}+\text{upper bound}}{2} \]

For example, for the class \(50 \leq t < 70\), the midpoint is

\[\begin{aligned} \text{midpoint} &= \frac{50+70}{2} \\ &= \frac{120}{2} \\ &= 60 \end{aligned}\]

So the point plotted would be \((60,\text{frequency})\).

F) When completing a frequency polygon, it is common to join the first and last plotted points down to the horizontal axis so the graph starts and ends at frequency \(0\). This helps show the shape of the distribution clearly.

G) Always read inequalities in class intervals carefully. For instance, \(70 \leq t < 90\) includes \(70\) but does not include \(90\). So a value of \(89.9\) belongs in this class, but a value of \(90\) belongs in the next class.