01 · Read a frequency
A table records x = 0, 1, 2 with frequencies 4, 3, 2. How many observations are there, and what is their total?
Hint
A zero-valued observation still counts.
Worked solution
Σf = 4 + 3 + 2 = 9. Σfx = 4×0 + 3×1 + 2×2 = 7.
Understand · explore · practise
Averages from frequency tables
Use frequencies to calculate exact means, locate medians and identify modes. Explore a table, solve unknown-frequency problems and check original worked practice.
Before you startMean, median and mode for listed data.
01 / Compress a list without losing counts
One row can stand for many observations.
Frequency f tells you how many times a value x occurs.
A table with four rows does not necessarily contain four observations.
Frequency tables efficiently store repeated values. For exact discrete values, the averages calculated from the table are the same as those from the expanded list.
Frequencies are repeated observationsExplore
| x | f | fx | Cumulative f |
|---|---|---|---|
| 0 | 2 | 0 | 2 |
| 1 | 3 | 3 | 5 |
| 2 | 2 | 4 | 7 |
| 3 | 1 | 3 | 8 |
Σfx = 10; Σf = 8; mean = 1.25.
Median = 1; mode = 1.
Expanded data: 0, 0, 1, 1, 1, 2, 2, 3.
The table is exact discrete data, not grouped intervals. A zero frequency contributes neither observations nor total.
02 / Count observations and total values
Keep Σf and Σfx distinct.
Values 1, 2, 4 have frequencies 3, 2, 5.Worked example
Σf = 3 + 2 + 5 = 10
There are ten observations.
Σfx = 3×1 + 2×2 + 5×4 = 27
This is the total of their values.
Mean = 27/10 = 2.7
Divide by observations, not rows.
02 · Wrong denominator
Why is (0 + 1 + 2)/3 not the mean of that dataset?
Hint
The frequencies are unequal.
Worked solution
It assigns equal weight to each distinct value instead of each observation. The actual mean is 7/9 ≈ 0.778.
03 / Calculate a weighted mean
Multiply first, then sum.
x̄ = Σfx / Σf
Each value is weighted by its frequency.
The mean must lie between the smallest and largest values that actually occur. A value shown with zero frequency is not an observation and should not define the observed extremes.
Watch: frequencies become repeated values
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Calculate the mean
Numbers of repairs are 0, 1, 2, 3 with frequencies 5, 4, 2, 1. Find the mean repairs per item.
Hint
There are 12 items; calculate the weighted total.
Worked solution
Total = 0 + 4 + 4 + 3 = 11 repairs. Mean = 11/12 ≈ 0.917 repairs per item.
04 · Include a zero frequency
Values 2, 4, 6 have frequencies 3, 0, 1. Find the mean.
Hint
The middle row adds zero to both sums.
Worked solution
Mean = (3×2 + 0×4 + 1×6)/(3 + 0 + 1) = 12/4 = 3.
04 / Use cumulative frequencies to locate ranks
Find values at the middle positions.
Put values in ascending order, add frequencies cumulatively and find which row contains the required middle rank or pair. The median is a data value or the average of two data values, not a frequency.
05 · Odd count
Values 1, 2, 3, 4 have frequencies 2, 4, 3, 2. Find the median.
Hint
The total is 11, so locate rank 6.
Worked solution
Cumulative frequencies are 2, 6, 9, 11. Rank 6 is value 2, so the median is 2.
06 · Even count across rows
Values 1, 2, 4 have frequencies 2, 3, 5. Find the median.
Hint
The total is ten; locate ranks 5 and 6 separately.
Worked solution
Cumulative frequencies are 2, 5, 10. Rank 5 is 2; rank 6 is 4. Median = (2 + 4)/2 = 3.
05 / Choose values with the greatest frequency
Do not report the frequency as the mode.
The mode is the value attached to the highest frequency. If the highest frequency is shared, report all corresponding modes. Check the variable’s units.
07 · Read the mode
Values 2, 5, 8 have frequencies 4, 9, 3. What is the mode?
Hint
Which value occurs nine times?
Worked solution
The mode is 5, not 9. Nine is the frequency of the modal value.
08 · Tied frequencies
Values 0, 1, 2, 3 have frequencies 2, 5, 5, 1. Find the modes.
Hint
There are two rows with maximum frequency.
Worked solution
1 and 2 are both modes; each occurs five times.
06 / Solve for a missing frequency
The unknown appears in the numerator and denominator.
Values 1, 3, 5 have frequencies 2, k, 4. The mean is 3.5.Worked example
(22 + 3k)/(6 + k) = 3.5
The weighted total and total count both depend on k.
22 + 3k = 21 + 3.5k
Multiply by the total frequency.
k = 2
Check non-negativity, integrality and the original mean.
09 · Find the frequency
Values 0, 2, 4 have frequencies 3, k, 2. Their mean is 1.75. Find k.
Hint
Use (2k + 8)/(k + 5) = 1.75.
Worked solution
2k + 8 = 1.75k + 8.75, so 0.25k = 0.75 and k = 3. Check: total 14 across 8 observations gives 1.75.
10 · An impossible stated mean
The same values 0, 2, 4 with frequencies 3, k, 2 are claimed to have mean 3. Is any non-negative frequency k possible?
Hint
Solve the equation, then check whether the result can be a count.
Worked solution
2k + 8 = 3(k + 5) gives k = −7, which cannot be a frequency. No non-negative k satisfies the claim.
07 / Check entries and interpretation
Values and frequencies are separate columns.
When using a calculator’s statistics table, enter the distinct values and their frequencies into the intended columns. Check the total observation count before trusting the summary. Device menus vary, but the arithmetic above provides a device-independent check.
11 · Diagnose a count
A calculator reports n = 3 for values 1, 2, 4 with frequencies 3, 2, 5. What likely went wrong?
Hint
The correct count is ten.
Worked solution
The frequencies were probably omitted or not enabled, so each distinct value was counted once. Check the entered frequency column and verify n = 10.
08 / A compact table still represents people or items
Translate the summary back to its units.
Check Σf, Σfx, middle ranks and the largest frequency. Expand a small table into a list when you need an independent check.
12 · Full summary
Values 0, 1, 2, 3 have frequencies 2, 3, 2, 1. Give mean, median and mode.
Hint
The expanded list is 0, 0, 1, 1, 1, 2, 2, 3.
Worked solution
Mean = 10/8 = 1.25. Middle ranks 4 and 5 both have value 1, so median 1. Value 1 has the largest frequency, so mode 1.
Section 1 of 8 · Compress a list without losing counts