FS4–FS5, HS5–HS6 - Averages from frequency tables

Mean from discrete frequencies

For values $x$ with frequencies $f$,

The product $fx$ accounts for all repetitions of value $x$. This is a calculation method to know, not a supplied formula.

Worked example 1: an exact discrete distribution

Question. Ten households give their number of bicycles as below. Find the mean, median, mode and range.

Plan. Add products for the mean, then use the running totals to locate the middle two of ten observations.

| Bicycles $x$ | Frequency $f$ | $fx$ | Running frequency total |
|---|---|---|---|---|
| $0$ | $2$ | $0$ | $2$ |
| $1$ | $4$ | $4$ | $6$ |
| $2$ | $3$ | $6$ | $9$ |
| $3$ | $1$ | $3$ | $10$ |
| Total | $10$ | $13$ | |

The mean is $13/10=1.3$. Positions $3$ through $6$ have value $1$, so both middle positions, $5$ and $6$, are $1$. The largest frequency is $4$, belonging to $1$ bicycle.

Answer. Mean $1.3$ bicycles, median $1$, mode $1$, range $3-0=3$.

Check. Expanding gives $0,0,1,1,1,1,2,2,2,3$, with total $13$ and ten values. The mean may be fractional even though each household has a whole-number count.

Worked example 2: different middle rows

Question. Find the four summaries for this distribution of books borrowed by ten pupils.

Plan. Keep the numerical values separate from their frequencies; locate positions $5$ and $6$.

| Books $x$ | Frequency $f$ | $fx$ | Running frequency total |
|---|---|---|---|---|
| $0$ | $0$ | $0$ | $0$ |
| $1$ | $2$ | $2$ | $2$ |
| $2$ | $3$ | $6$ | $5$ |
| $4$ | $4$ | $16$ | $9$ |
| $5$ | $1$ | $5$ | $10$ |
| Total | $10$ | $29$ | |

Mean $=29/10=2.9$. Position $5$ is $2$; position $6$ is $4$, giving median $(2+4)/2=3$. The most frequent value is $4$.

Answer. Mean $2.9$, median $3$, mode $4$, range $5-1=4$ books.

Check. The expanded list is $1,1,2,2,2,4,4,4,4,5$. No pupil borrowed zero books, so zero must not be used as the observed minimum.

Watch out

Do not average the row labels or the frequencies. Divide the total value, found from the products, by the total number of observations. Repeated values count repeatedly.

Modal and median classes

For the equal-width groups here, the modal class is the interval with the greatest frequency. Locate the middle observation or observations using running frequency totals. If the middle observations lie in one interval, that interval is the median class. Neither class gives an exact individual mode or median.

Worked example 3: grouped mean and classes

Question. Twenty journey times are grouped below. Estimate the mean and identify the modal and median classes.

Plan. Use midpoints for an estimated total, and the actual frequencies for the class positions.

The greatest frequency is $10$, in $10\le t<20$. The middle positions are $10$ and $11$; both are among positions $5$ to $14$, in that same class.

Answer. Estimated mean $16$ minutes; modal class and median class both $10\le t<20$.

Check. Frequencies total $20$. A midpoint-only mean $(5+15+25)/3=15$ would give the groups equal weight despite their different counts.