Find measures of centre from a frequency display

Twenty households have the pet counts shown. What is the mean pets per household?

Grow the last frequency from 0 to 5 and 10. Compare the observation count, weighted total and mean with the unweighted average of the displayed labels

For every position of the slider, what happened as the last bar grew from 0 to 10?

Read values and their weights

A frequency counts repeated observations. A value label is not a frequency. The mean uses the weighted total; the median uses central positions in the expanded ordered data

1. 1Count Add frequencies, including the frequencies attached to zero values \(\displaystyle n=\sum f\)
2. 2Mean Multiply each value by its frequency, add the products, then divide by the count \(\displaystyle \bar{x}=\frac{\sum xf}{n}\)
3. 3Median For an even count, locate both central positions and average their values. A zero-frequency row contributes no observations

How many matches are represented by this goals display?

How many goals were scored altogether?

Find the mean goals per match

Which ordered position locates the median of these 25 matches?

Now report the median goals per match, not its position

This table represents six observations. Which positions determine its median?

Find the median of this frequency table

Find the mean of this table. An exact fraction is accepted

What does the zero frequency beside value 1 mean?

Scale the pattern from 1 to 2 and 4 copies. Watch the counts and total change while the exact percentages stay fixed

What stays fixed when every frequency is multiplied by the same positive number?

The total count is now hidden. Can this exact percentage table still determine the mean?

Calculate the mean using only these exact percentages

What is the median? The cumulative percentages are 20%, 60%, 100%

Exact weights and rounded weights

Exact percentage weights determine a mean without an absolute count. Rounded percentages can hide the true proportions, so a calculation using them is generally an estimate

1. 1Do not invent a sample size A percentage table may fit several different counts. Work with its proportions directly
2. 2Check the central boundary If cumulative percentage is exactly 50% between two distinct values, an even-sized dataset needs both central values. Do not automatically take only the next row

Eleven whole-number scores were grouped as shown. Which interval contains the median?

Can the exact median score be identified from those intervals alone?

Using class midpoints gives mean 24.5. How should this result be described?

The same grouped counts could represent three tens, five twenties and three thirties, or three nineteens, five twenty-nines and three thirty-nines. What are the two possible means?

Values 1, 3 and 7 have frequencies 2, 1 and 3. What is the mode?

Explain what is recoverable

Complete a frequency-to-percentage and grouping investigation using this exact table: values 0, 2, 6, 8 have counts 2, 3, 4, 1 respectively. Produce the count, weighted total, mean, cumulative counts and median, then the exact percentages and percentage-based mean. Explain every denominator and both central positions

1. 1Compare information loss Group the values into intervals. Explain which exact values are now hidden and why midpoint calculations need an estimate label
2. 2Independent explanation Write your reasoning and check it yourself or with a teacher. A correct choice alone does not demonstrate a complete written explanation
3. 3Choose and use technology Choose suitable statistical functions or spreadsheet formulas for a frequency table. Explain your choice and record the value/frequency inputs, formulas or settings and outputs. Verify weighted totals and central positions independently from the expanded ordered list. Explain why applying AVERAGE or MEDIAN to the four value labels alone is generally wrong
4. 4Group and test information loss Use the intervals from 0 inclusive to 4 exclusive, from 4 inclusive to 8 exclusive, and from 8 inclusive to 12 exclusive. Produce their frequencies and midpoint estimate of the mean. Locate both central observations by class. Then construct a different list of ten non-negative whole numbers with the same grouped frequencies but a different exact mean and median. Use it to explain why the grouped table alone does not determine either exact summary
5. 5Complete before continuing Save the complete calculations, second dataset, explanation and reproducible tool record before opening the model. Writing, dataset construction and external-tool execution are not automatically marked; use self or teacher review

Model response: exact values, weights and grouping limits

The exact value-frequency table represents ten records with total 38, mean 3.8 and median 4. Grouping these values loses information: its midpoint mean is an estimate, and the middle pair lie in different classes

1. 1Exact table and middle pair Expand the original table as 0, 0, 2, 2, 2, 6, 6, 6, 6, 8. The count is 2+3+4+1=10 and the weighted total is 0×2+2×3+6×4+8×1=38. The mean is 38/10=3.8. Cumulative counts 2, 5, 9, 10 place the fifth record at 2 and sixth at 6, giving median (2+6)/2=4. A zero value contributes observations even though it contributes zero to the total
2. 2One reproducible tool record Put values in A2:A5 and frequencies in B2:B5. SUM(B2:B5) gives 10 and SUMPRODUCT(A2:A5,B2:B5) gives 38; divide the latter by the former for 3.8. In C2 use SUM($B$2:B2) and fill down to obtain cumulative frequencies. Locate positions 5 and 6 at the first cumulative counts reaching them, 5 and 9, then average their associated values 2 and 6. These formulas use every record’s frequency; AVERAGE(A2:A5) would give the incorrect unweighted mean 4.
3. 3Exact percentages Use 100×frequency/10 to obtain 20%, 30%, 40%, 10%. The mean is (0×20+2×30+6×40+8×10)/100=3.8. Exact percentages preserve relative weights; an absolute count is not required for this mean. The cumulative percentages 20%, 50%, 90%, 100% put the middle pair across the 50% boundary, giving median (2+6)/2=4. These percentages are exact, unlike rounded percentages that may hide the true weights
4. 4Grouped estimate and central classes The specified classes have frequencies 5, 4, 1. The value 8 enters the class starting at 8, not the class ending there. Midpoints 2, 6, 10 give estimated mean (2×5+6×4+10×1)/10=4.4. Positions 5 and 6 lie in the first and second classes respectively. The grouped table does not reveal either actual central value. We know the original median only because the ungrouped table was retained
5. 5A compatible counterexample A second list is 1, 1, 3, 3, 3, 7, 7, 7, 7, 11. It has the same class frequencies 5, 4, 1, but total 50, mean 5 and median (3+7)/2=5. Its grouped midpoint estimate is still 4.4. Two compatible lists with different exact summaries show why that estimate and a central class cannot determine the exact original summaries
6. 6Independent checks and self-review Compare every frequency with the expanded lists, both sample counts, weighted totals, central positions, exact percentages, interval membership and counterexample. Check that the grouped midpoint result is labelled an estimate and that the tool record distinguishes raw data from frequency inputs. Different valid counterexamples and tools are welcome; their complete construction and execution remain self or teacher reviewed

Which exact summaries follow from this ungrouped frequency table?