Mean and Mode
Theory
The mean \(\bar{x}\) is the sum of the values divided by the number of values; the mode is the value that appears most often. Turn the formula around, \(\text{total} = \text{mean} \times \text{number of values}\), to answer backwards questions. This Year 8 maths page shows how.
The mean. \(\bar{x} = \dfrac{\text{sum of the values}}{\text{number of values}}\). Values and means may be negative or decimal, and the mean need not be one of the values.
The mode is the value (or values) with the highest frequency. A set where every value appears equally often has no mode; one mode is unimodal, two modes bimodal.
From a frequency table or a dot plot. Multiply each value by its frequency and add, to get the total. Add the frequencies, to get the number of values. The mode is the value with the largest frequency (the tallest column of dots).
Turn the formula around: \(\text{total} = \text{mean} \times \text{number of values}\). This answers every backwards question: a missing value, the score needed for a target mean, or the new mean after a value is removed. If a wrong value was entered, add the difference to the total and divide again.
Adding or removing one value. A value above the mean pulls the mean up; a value below it pulls it down; a value equal to the mean changes nothing. An outlier, a value far from the rest, can pull the mean a long way while the mode stays put.
Two groups. Add the two totals and add the two counts, then divide. Averaging the two means works only when the groups are the same size.
Naming the modes:
| Data | Modes | Name |
|---|---|---|
| \(3,~8,~12,~15\) | none | no mode |
| \(6,~9,~9,~9,~14\) | \(9\) | unimodal |
| \(4,~7,~7,~10,~13,~13\) | \(7\) and \(13\) | bimodal |
Mean from a frequency table
- Multiply each value by its frequency.
- Add the products for the total, and add the frequencies for the number of values.
- Divide the total by the number of values.
Backwards questions
- Total needed \(=\) mean \(\times\) number of values.
- Compare with the total you know; the difference is the missing value (or the change).
| \(\dfrac{122 + x}{9}\) | \(=\) | \(15\) |
| \(122 + x\) | \(=\) | \(9 \times 15 = 135\) |
| \(x\) | \(=\) | \(13\) |
\(x = 13\)
Without \(x\), the values \(12\), \(15\) and \(16\) each appear twice.
For \(16\) to occur most often, \(x = 16\).
| \(15 + 42 + 42\) | \(=\) | \(99\) |
| \(99 + 40 + 36\) | \(=\) | \(175\) |
| \(\text{Students}\) | \(=\) | \(3 + 7 + 6 + 5 + 4\) |
| \(\) | \(=\) | \(25\) |
| \(\text{Mean}\) | \(=\) | \(\dfrac{175}{25} = 7\) |
Mean \(= 7\). The largest frequency is \(7\), for a mark of \(6\), so the mode is \(6\).
| \(34 \times 12\) | \(=\) | \(408\) |
| \(39 \times 18\) | \(=\) | \(702\) |
| \(\dfrac{408 + 702}{30}\) | \(=\) | \(\dfrac{1110}{30} = 37\) |
\(37\) seconds. Averaging the means gives \(36.5\) s, which is wrong: the slower heat has more swimmers.
(i) A \(20\)th player scores \(82\). Find the new mean.
(ii) What would that player have needed for a mean of \(64\)?
| \(62 \times 19\) | \(=\) | \(1178\) |
| \(1178 + 82\) | \(=\) | \(1260\) |
| \(1260 \div 20\) | \(=\) | \(63\) |
| \(64 \times 20\) | \(=\) | \(1280\) |
| \(1280 - 1178\) | \(=\) | \(102\) |
(i) \(63\) points (\(82\) is above the mean, so the mean rises)
(ii) \(102\) points
(ii) \(20\) students had a mean of \(70\), but one mark was entered as \(64\) instead of \(46\). Find the correct mean.
(iii) Eight values have a mean of \(5\). The value \(12\) is removed. Find the new mean.
| \(-3 + 1 - 5 + 2 + 0\) | \(=\) | \(-5\) |
| \(-5 \div 5\) | \(=\) | \(-1\) |
| \(1400 - 64 + 46\) | \(=\) | \(1382\) |
| \(1382 \div 20\) | \(=\) | \(69.1\) |
| \(40 - 12\) | \(=\) | \(28\) |
| \(28 \div 7\) | \(=\) | \(4\) |
(i) \(-1\) degree
(ii) \(69.1\)
(iii) \(4\) (removing a value above the mean pulls the mean down)
| \(\dfrac{34 + 2f}{12 + f}\) | \(=\) | \(2.5\) |
| \(34 + 2f\) | \(=\) | \(30 + 2.5f\) |
| \(4\) | \(=\) | \(0.5f\) |
| \(f\) | \(=\) | \(8\) |
\(f = 8\). Check: total \(50\) over \(20\) players is \(2.5\).
Common pitfalls
Frequently asked questions
How do you find the mean?
Add all the values and divide by how many there are: the mean of \(4\), \(6\), \(11\) is \(21 \div 3 = 7\).
How do you find the mean from a frequency table?
Multiply each value by its frequency, add these products, then divide by the total frequency.
Can there be more than one mode?
Yes. Two modes is bimodal; if every value appears equally often there is no mode.
How do I find a missing value from the mean?
Work out the total needed (mean \(\times\) number of values) and subtract the values you know.
How does an outlier affect the mean?
It pulls the mean towards it, sometimes a long way, while the mode usually stays the same.