Exercise solution 7, page 66 - Mathematics Higher Student Book

⭐ What you need to know:

A) A grouped frequency table shows how many data values fall inside each interval. Here, the heights are grouped into class intervals such as \(150 \leq h < 160\).

This means:

\(150 \leq h < 160\) includes values from \(150\) up to, but not including, \(160\).

For example, if a class is \(120 \leq x < 130\), then \(120\), \(124\) and \(129.9\) are included, but \(130\) is not included.

B) In a frequency diagram for grouped data with equal class widths, each bar covers one interval on the horizontal axis, and the height of the bar is the frequency.

If all intervals have the same width, the taller the bar, the greater the frequency. So to complete the diagram, you read each frequency from the table and draw a bar up to that value.

For example, if the intervals are \(40 \leq m < 50\), \(50 \leq m < 60\), \(60 \leq m < 70\) with frequencies \(4\), \(9\), and \(7\), then the three bars should have heights \(4\), \(9\), and \(7\).

C) 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 class boundaries and divide by \(2\):

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

For example, for the class \(30 \leq y < 40\):

\[ \begin{aligned} \text{midpoint} &= \frac{30+40}{2} \\ &= \frac{70}{2} \\ &= 35 \end{aligned} \]

So the point for this class would be plotted at \((35,\text{frequency})\).

D) When drawing a frequency polygon, it is important to use the centre of each class, not the class boundary.

For example, if the class is \(80 \leq t < 90\) and the frequency is \(12\), you plot the point at \(t=85\), not at \(80\) or \(90\). So the plotted point is \((85,12)\).

After plotting all the midpoints with their frequencies, join the points in order using straight line segments.

E) If the first or last class has frequency \(0\), this helps the frequency polygon start and finish on the horizontal axis.

For example, if the class \(10 \leq p < 20\) has frequency \(0\), its midpoint is:

\[ \begin{aligned} \frac{10+20}{2} &= \frac{30}{2} \\ &= 15 \end{aligned} \]

So the point \((15,0)\) is plotted on the axis. This makes the polygon begin or end neatly.

F) If all the class widths are equal, a quicker way to draw the frequency polygon is to plot the midpoints directly from the table, without drawing the full bar chart first.

The method is:

find each class midpoint,

plot each midpoint against its frequency,

join the points with straight lines.

This is often faster because you do not need to draw all the bars first.

G) Always check that the total frequency matches the total number of data values given in the question.

For grouped data, this means adding all the frequencies in the table.

For example, if the frequencies are \(3\), \(8\), \(11\), \(14\), and \(4\), then:

\[ \begin{aligned} 3+8+11+14+4 &= 11+11+14+4 \\ &= 22+14+4 \\ &= 36+4 \\ &= 40 \end{aligned} \]

This tells you there are \(40\) values altogether. Checking the total helps you spot mistakes in the table or diagram.