E9.3 - Grouped-data averages

E9.3 - Grouped-data averages

A grouped table tells you which interval contains each observation, but it hides the exact values. You will use midpoints to estimate a mean, explain why it is an estimate, and identify the modal class without pretending to know an exact mode.

Class intervals and midpoints

Can you recover an exact mean when you know only how many observations lie between ten and twenty? Usually not: the observations could be near either end. A midpoint gives one representative value for the interval.

A class interval groups a range of values. Read its endpoints carefully: 10<t≤2010<t\le20 includes twenty but excludes ten, while 10≤t<2010\le t<20 does the reverse. Adjacent classes must place a boundary observation in one class, not two.

Class midpoint

midpoint=lower endpoint+upper endpoint2\text{midpoint}=\frac{\text{lower endpoint}+\text{upper endpoint}}{2}

It is halfway between the endpoints. For grouped discrete integers 10,11,…,1910,11,\ldots,19, use (10+19)÷2=14.5(10+19)\div2=14.5. A representative midpoint need not itself be a possible observed value. This formula is not given in the List of formulas: learn it.

For a continuous interval from ten to twenty, the midpoint is fifteen regardless of which endpoint is included. It is a representative value, not a claim that every observation equals fifteen or that the observations are evenly spaced.

Worked example 1: continuous intervals

Question. Find the midpoints of these time classes: 0<t≤100<t\le10, 10<t≤2010<t\le20 and 20<t≤3020<t\le30, with tt in minutes. Which class contains exactly ten minutes?

Plan. Average each pair of endpoints and use the inequality symbols for the boundary observation.

First midpoint=0+102=5halfway from zero to tenSecond midpoint=10+202=15halfway from ten to twentyThird midpoint=20+302=25halfway from twenty to thirty\begin{aligned} \text{First midpoint}&=\frac{0+10}{2}=5 && \text{halfway from zero to ten}\\ \text{Second midpoint}&=\frac{10+20}{2}=15 && \text{halfway from ten to twenty}\\ \text{Third midpoint}&=\frac{20+30}{2}=25 && \text{halfway from twenty to thirty} \end{aligned}

Answer. Midpoints are five, fifteen and twenty-five minutes. Ten minutes belongs to 0<t≤100<t\le10, because its upper endpoint is included.

Check. Each midpoint is five minutes from either endpoint of its class. The second class requires more than ten, so ten has not been counted twice.

Change the variable to a count of items. The first class is now the whole numbers zero to nine inclusive, rather than a continuous interval ending at ten.

Worked example 2: grouped discrete values

Question. Find the midpoints of the integer-count classes 00–99, 1010–1919 and 2020–2929. Which class contains nineteen?

Plan. Average the smallest and largest integers in each class.

First midpoint=0+92=4.5endpoints are zero and nineSecond midpoint=10+192=14.5endpoints are ten and nineteenThird midpoint=20+292=24.5endpoints are twenty and twenty-nine\begin{aligned} \text{First midpoint}&=\frac{0+9}{2}=4.5 && \text{endpoints are zero and nine}\\ \text{Second midpoint}&=\frac{10+19}{2}=14.5 && \text{endpoints are ten and nineteen}\\ \text{Third midpoint}&=\frac{20+29}{2}=24.5 && \text{endpoints are twenty and twenty-nine} \end{aligned}

Answer. The midpoints are 4.5,14.5,24.54.5,14.5,24.5 items. Nineteen belongs to the 1010–1919 class.

Check. Zero and nine are equally far from 4.54.5; ten and nineteen are equally far from 14.514.5. A midpoint of five for the first class would be nearer nine than zero, so it would not be halfway.

Both examples use the centre of the stated interval. The midpoint represents its class later in a calculation; it does not replace its boundaries when deciding which class contains an observation.

Watch out

The midpoint of 10<t≤2010<t\le20 is fifteen; the midpoint of the integer class 1010–1919 is 14.514.5. Use the endpoints actually given, rather than automatically adding half of ten to a lower label.

Read a new pair of adjoining classes.

Estimating a grouped mean

Recall the exact-value frequency-table mean: multiply each value by its frequency, add, then divide by total frequency. In grouped data, replace the unknown individual values by their class midpoint. This estimates the total, so it estimates the mean.

Estimated grouped mean

For nonempty grouped numerical data with class midpoints mm and frequencies ff,

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

The denominator counts observations, not classes. This formula is not given in the List of formulas: learn it.

The estimate treats each class's observations as if their average were its midpoint. The actual class average may be different. Two raw lists can therefore give the same grouped table and different exact means.

Worked example 3: grouped continuous measurements

Question. Estimate the mean time in this table.

Time tt, minutesFrequency ffMidpoint mmfmfm
0<t≤100<t\le1044552020
10<t≤2010<t\le20881515120120
20<t≤3020<t\le303325257575

Plan. Estimate each class total using midpoint times frequency, then divide the estimated total by fifteen observations.

Estimated total=20+120+75=215 minadd the class estimatesTotal frequency=4+8+3=15count observationsEstimated mean=21515=14.3 min (3 s.f.)divide total by count\begin{aligned} \text{Estimated total}&=20+120+75=215\,\mathrm{min} && \text{add the class estimates}\\ \text{Total frequency}&=4+8+3=15 && \text{count observations}\\ \text{Estimated mean}&=\frac{215}{15}=14.3\,\mathrm{min}\ (3\text{ s.f.}) && \text{divide total by count} \end{aligned}

Answer. The estimated mean is 14.314.3 minutes, to three significant figures.

Check. Starting from fifteen minutes, the four first-class representatives are ten below and the three last-class representatives are ten above. Their net shortfall is ten minutes, so the mean is 15−10/15=14.333…15-10/15=14.333\ldots, confirming the estimate independently.

Keep the frequencies but replace the classes by integer counts zero to nine, ten to nineteen and twenty to twenty-nine. Every representative value falls by one half.

Continue this lesson.

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

Create free account