6.2 - Grouped mean estimates and modal class

Grouping measurements makes a table shorter but hides their precise values. Identify a modal class and estimate a mean with class midpoints, while explaining what information has been lost.

Classes and the modal class

Can the same grouped table fit one dataset with mean $6.5$ and another with mean $13.5$ ? First, learn exactly what a class tells you.

You already know

A frequency counts observations. For a discrete frequency table, multiply each exact value by its frequency, add the products and divide by the total frequency to find a mean.

A grouped table gives intervals rather than individual values. Read its inequalities: $10\leq d<20$ includes $10$ but excludes $20$ . An observation of $20$ belongs in the next class starting at $20$ , so it is counted once.

Modal class

The class interval with the greatest frequency. Give the interval, not its frequency or midpoint. If classes tie for greatest frequency, identify both.

The table usually does not reveal which exact individual value occurs most often. The modal class describes a group of values, not a known raw-data mode. Our modal-class examples use equal-width classes.

Worked example 1: identify the most frequent class

Question. Find the modal class for journey distances.

Distance $d$ in kilometres	Frequency
$0\leq d<10$	$4$
$10\leq d<20$	$9$
$20\leq d<30$	$7$

Plan. Find the greatest frequency and report its interval.

\begin{aligned} \text{greatest frequency}&=9 &&\text{compare 4, 9 and 7}\\ \text{corresponding class}&:10\leq d<20 &&\text{read its row label} \end{aligned}

Why report an interval? The $9$ journeys may have different distances within that class.

Answer. $10\leq d<20$ .

Check. Its count $9$ exceeds both other counts; the table has $4+9+7=20$ journeys.

Change only the second frequency from $9$ to $4$ . The interval boundaries do not change.

Worked example 2: change one class frequency

Question. Find the revised modal class.

Plan. Compare the new frequencies before reading the row label.

\begin{aligned} \text{frequencies}&:4,4,7 &&\text{replace the second count}\\ \text{greatest frequency}&=7 &&\text{third row is now greatest}\\ \text{modal class}&:20\leq d<30 &&\text{report its interval} \end{aligned}

Why is it not the second class anymore? Modal class depends on the counts, not the numerical size of the boundaries.

Answer. $20\leq d<30$ .

Check. The first two classes tie at $4$ , but neither reaches the maximum $7$ .

A midpoint is useful for the mean method in the next part. It is halfway between the lower and upper boundaries. Here the three midpoints are $(0+10)/2=5$ , $(10+20)/2=15$ and $(20+30)/2=25$ kilometres.

Watch out

The line $\text{modal class}=9$ gives a frequency. For the first table the answer is $10\leq d<20$ , not $9$ and not its midpoint $15$ .

Pearson's “Write down the modal class” asks for that full interval. Copy its endpoint signs accurately.

Estimating the grouped mean

Replace all unknown values in each class by its midpoint to produce an estimated total and mean.

This replacement uses one central representative, not a claim that every actual value equals it. If midpoint $m$ represents $f$ observations, the class contributes approximately $fm$ to the total. Classes with more observations must have more weight.

Grouped mean estimate

m=\frac{\text{lower boundary}+\text{upper boundary}}{2}

\text{estimated mean}=\frac{\sum fm}{\sum f}

The symbol $\sum$ means add all entries. Not in the Foundation or Higher formulae sheet: learn these relationships.

Use a midpoint column and a product column. The estimated mean keeps the measured variable's unit, not the frequency unit.

Worked example 3: weight the midpoints

Question. Estimate the mean of these times.

Time $t$ in minutes	Frequency $f$	Midpoint $m$	Product $fm$
$0\leq t<10$	$3$	$5$	$15$
$10\leq t<20$	$5$	$15$	$75$
$20\leq t<30$	$2$	$25$	$50$

Plan. Add midpoint products and divide by total frequency.

\begin{aligned} \sum f&=3+5+2=10 &&\text{count all observations}\\ \sum fm&=3\times5+5\times15+2\times25=140 &&\text{estimated total minutes}\\ \text{estimated mean}&=140\div10=14 &&\text{estimated total per observation} \end{aligned}

Why multiply by frequency? The midpoint $15$ stands for five observations, not one.

Answer. The estimated mean is $14\,\mathrm{min}$ .

Check. It lies between the representative values $5$ and $25$ ; multiplying by $10$ recovers the estimated total $140$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Number

Algebra

Shape, space and measure

Handling data

6.2 - Grouped mean estimates and modal class

Classes and the modal class

Estimating the grouped mean

Continue this lesson.

Learn faster with Nova